Motion in a Straight Line

Motion in a Straight Line INTRODUCTION The branch of physics which deals with the study of motion of material objects is called mechanics. Mechanics is divided into following branches. (i) Statics : Statics is the branch of mechanics which deals with the study of motion of objects under the effect of forces in equilibrium. (ii) Kinematics : It is that branch of mechanics which deals with the study of motion of object without taking into account the factors (i.e. nature of forces, nature of bodies etc.) which cause motion. Here time factor plays an essential role. (iii) Dynamics : It is that branch of mechanics which deals with the study of motion of objects taking into account the factors which cause motion. Rest : An object is said to be at rest if it does not change its position with time, with respect to its surroudings. A book lying on a table, a person sitting in a chair are the examples of rest. Motion : An object is said to be in motion if it changes its position with time, with respect to its moving on rails, a ship sailing on water, a man walking on road are some of the examples of motion, visible to the eye. Motion of gas molecules is an example of motion, invisible to the eye. Rest & Motion are relative terms : When we say that an object is at rest or in motion,then this statement is incomplete and meaningless. Basically, rest & motion are relative terms. An object which is at rest can also be in motion simultaneously. This can be illustrated as follows. The passengers sitting in a moving bus are at rest with respect to each other but they are also in motion at the same time with respect to the objects like trees, buildings on the road side. So the motion and rest are relative terms. Rectilinear motion : If a particle moves in a fixed direction, the motion of this type is called rectilinear motion or one dimensional motion. For example the motion of an ant on a wire is a rectilinear motion. Motion in a Straight Line Two dimensional motion : If the motion of a particle is in such a way that its position remains on a fixed plane, then the motion of a particle is called two dimensional motion. For example the motion of a rolling ball on a horizontal plane or earth’s surface, is a two dimensional motion. Three dimensional motion : If a line or plane is not fixed for the motion of a particle, then its motion is called three dimensional motion. DISPLACEMENT The shortest distance from the initial position to the final position of the particle is called displacement. The displacement of a particle is measured as the change in the position of the particle in a particular direction over a given time interval. It depends only on final & initial positions. Displacement of a particle is a position vector of its final position w.r.t. initial position. For example the motion of a flying bird is a three dimensional motion. DISTANCE The length of the actual path between initial & final positions of a particle in a given interval of time is called distance covered by the particle. Distance is the actual length of the path. It is the characteristic property of any path ie., path is always associated when we consider distance between two positions. Distance beteen A & B while moving through path (1) may or may not be equal to the distance between A & B while moving through path (2)  Position vector of A w.r.t. O  OA   rA  x 1 î  y1 ĵ  z1 k̂  Position vector of B w.r.t. O  OB   rB  x 2 ˆi  y 2 ˆj  z2kˆ Displacement   AB  x 2  x1 ˆi  y 2  y1 ˆj  z 2  z1 kˆ Characteristics of Distance : Characteristics of Displacement : (i) It is a scalar quantity. (i) It is a vector quantity. (ii) It depends on the path. (iii) It never reduces with time. (ii) The displacement of a particle between any two points is equal to the shortest distance between them. (iv) Distance covered by a particle is always positive & can never be negative or zero. (iii) The displacement of an object in a given time interval can be positive, negative or zero. (v) Dimension :  M0 L1 T 0  (iv) Dimension : M0L1T0  (vi) Unit : In C.G.S.system centimeter (cm), In S.I. system metre (m). (v) Unit : In C.G.S. centimeter (cm), In S.I. system meter (m). COMPARATIVE STUDY OF DISPLACEMENT AND DISTANCE S.No. Displacement Distance 1. It has single value bewteen two points It may have more than one value between two points 2. 0  Displacement  0 Distance > 0 3. Displacement can decrease with time It can never decrease with time. 4. It is a vector quantity It is a scalar quantity Motion in a Straight Line Special note: Ex.4 A monkey is moving on circular path of radius 80 m. Calcualte the distance covered by the 1. The actual disatnce travelled by a particle in the monkey. given interval of time is always equal to or greater than the magnitude of the displacement and in no Sol. Distance = Circumference of the circle case, it is less than the magnitude of the displacement, D  2R  D  2  80  160  3.14  502.40m i.e., Distance  | Displacement | Ex.5 A man has to go 50 m due north, 40 m due east 2. Displacement may be + ve, – ve or zero. and 20 m due south to reach a field. 3. Distance, speed and time can never be negative. (a) What distance he has to walk to reach the field? 4. At the same time particle cannot have two (b) What is his displacement from his house to the positions. field? Some Impossible graphs : Sol. Let origin be O then (a) Distance covered = OA + AB + BC = 50 + 40 + 20 = 110 meter (b) First method : Displacement OC = Ex.1 An old person moves on a semi circular track of radius 40 m during a morning walk. If he starts at one end of the track and reaches at the other end. Find the displacement of the person. = OD2  CO 2 = 10 25  50 meter 402  30 2 Second method : Displacement  d  50ˆj  40iˆ  20 ˆj = 30 ˆj  40 ˆi | d |  40 2  30 2 = 50 meter Sol. Displacement = 2R = 2 x 40 = 80 meter.  1 Ex.2 An athelete is running on a circular track of radius Ex.6 A body covers 4 th part of a circular path. Calulate the ratio of distance and displacement. 50 meter. Calculate the displacement of the athlete after completing 5 rounds of the track. Sol. Since final and initial positions are same . Hence displacement of athlete will be r  r  r  0 Ex.3 If a particle moves from point A to B then distance covered by particle will be. Sol. D = x + 2x = 3x Sol. Distance = AB from path (1) = Displacement = AB = = OA 2  OB 2 r2  r2  r 2 Distance r / 2     Displacement r 2 2 2 2r r  4 2 Motion in a Straight Line Ex.7 A point P consider at contact point of a wheel on ground which rolls on ground without sliping then value of displacement of point P when wheel completes half of rotation - [If radius of wheel is 1 m] Sol. Displacement = 2  4 Ans. speed= distance travelled time taken (i) It is a scalar quantity. (ii) It gives no idea about the direction of motion of the object. (iii) It can be zero or positive but never negative. (iv) Unit: C.G.S. cm/sec, S.I. m/sec, 100 5  m/ s 60  60 18 5 1 km / h  m/s 18 (v) Dimension : M0 L1 T 1 1km / h = SPEED Speed of an object is defined as the time rate of change of position of the object in any direction. It is measured by the distance travelled by the object in unit time in any direction. i.e., Types of speed : (a)Uniform speed : An object is said to be moving with a uniform speed, if it covers equal distances in equal intervals of time, how soever small these intervals may be. The uiform speed is shown by straight line in distance time graph. For example: suppose a train travels 1000 metre in 60 second. The train is said to be moving with uniform speed, if it travels 500 metre in 30 second, 250 metre in 15 second, 125 metre in 7.5 second and so on. (c) Average Speed : When an object is moving with a variable speed, then the average speed of the object is that constant speed with which the object covers the same distance in a given time as it does while moving with variable speed during the given time. Average speed for the given motion is defined as the ratio of the total distance travelled by the object to the total time take i.e., Average speed V = total distance travelled total time taken (b)Non Uniform Speed : An object is said to be moving with a variable speed if it covers equal Note : If any car covers distance x1, x2, ........... in the distance in unequal intervals of time or unequal time intervals t1, t2, ....... then. distances in equal intervals of time, howsoever small x  x 2  x3  .......  xn V= 1 these intervals may be. t  t  .....  t 1 For example : suppose a train travels first 1000 metre in 60 second, next 1000 metre in 120 second and next 1000 metre in 50 second, then the train is moving with variable speed. 2 n Motion in a Straight Line SOME IMPORTANT CASES RELATED TO AVERAGE SPEED : Case : 1 When t  0 , then average speed of the body becomes the instantaneous speed. Instantaneous speed =Lim t  0 If body covers distances x1, x2, and x3 with speeds v1, v2, and v3 respectively in same direction then average speed of body. V  x dx  t dt Note : (a) Speedometer of the vehicle measures its instantaneous speed. x1  x 2  x 3 t1  t 2  t 3 (b) In case of a uniform motion of an object, the instantaneous speed is equal to its uniform speed. x1  x 2  x 3 x2 x3 V  x x1 x x 1 here, t1  v ,t 2  v ,t 3  v  2  3 1 2 3 v1 v 2 v 3 VELOCITY It is defined as rate of change of displacement. Characteristics of Velocity : (i) It is a vector quantity. If body covers equal distances with different speeds (ii) Its direction is same as that of displacement. then, x1  x 2  x3  x (iii) Unit and dimension : Same as that of speed. 3v1v2v3 3x 3 V   x1 x x 1 1 1 v1v2  v2v3  v3v1     v1 v2 v3 v1 v2 v3 Case : 2 If any body travels with speeds v1,v2,v3 during time intervals t1,t2,t3 respectively then the average speed of the body wil be Average speed V x1  x 2  x 3 v1t1  v 2 t 2  v 3 t3  t1  t 2  t 3 t1  t 2  t 3 If t1  t 2  t3  t  v1  v 2  v 3  t 3t Types of Velcoity : (a) Instantaneous Velocity (b) Average Velocity (c) Uniform Velocity (d) Non-uniform Velocity (a)Instantaneous Velocity : It is defined as the velocity at some particular instant.    r dr  Instantaneous velocity  lim t 0 t dt (b)Average Velocity : Average Velocity =  v1  v 2  v 3 3 (d)Instantaneous speed : The speed of the body at any instant of time or at a particular position is called instantaneous speed. Let a body travel a distance  x in the time interval  t , then its average speed  x . t Total Displacement Total time (c) Uniform Velocity : A particle is said to have uniform velocity, if magnitudes as well as direction of its velocity remains same and this is possible only when the particles moves in same straight line reservering its direction. (d)Non-uniform Velocity : A particle is said to have non-uniform velocity, if either of magnitude or direction of velocity changes (or both changes). Motion in a Straight Line COMPARATIVE STUDY OF SPEED AND VELOCITY Ex.8 A man walks at a speed of 6 km/hr for 1 km and 8 km/hr for the next 1 km. What is his average speed for the walk of 2 km. 2v v Ex.9 The distance travelled by a particle S = 10t2 (m). Find the value of instantaneous speed at t = 2 sec. 0 2t  20t Therefore,  268 1 2 Sol. V= v  v  6  8  7 km / h 1 2 Sol. V  (ii) Total displacement = zero, since the car returns to the original position. dx d  10t 2  10 2t  20t dt dt average velocity total displacement 0  0 time taken 2t Ex.11 From the adjoining position time graph for two particles A and B the ratio of velocities vA : vB will be Put t  2 sec . V  20  2  40m / s Put t  2 sec . V  20  2  40 m / s Ex.10 A car travels a distance A to B at a speed of 40 km/h and returns to A at a speed of 30 km/h. (i) What is the average speed for the whole journey? [1] 1 : 2 [2] 1: 3 (ii) What is the average velocity ? V tan  tan 300 a   Sol. (i) Let AB = s, time taken to go from A to B, t1 Sol. A  VB tan b tan 600 10 2t  20t = s s h and time taken to go from B to A, t2 = h 40 30  total time taken = t 1 + t2 = 3  4 s 7s s s    h 40 30 120 120 Total distance travelled = s + s = 2s  Average speed dx d V  10t 2  10 2t  20t dt dt Put t  2 sec . V  20  2  40m / s [3] 1/ 3 3 3 :1  1 3 [4] 1 : 3 Ans. [4] Ex.12 The position of a particle moving on x-axis is given by x = At3 + Bt2 + Ct + D. The numerical value of A, B, C, D are 1, 4, –2 and 5 respectivley and S.I. units are used. Find velocity of the particle at t = 4 sec. Sol. V  dx d  dt dt [At3 + Bt2 + Ct + D] or V = 3At2 + 2Bt + C Put t  2 sec . V  20  2  40 m/s V = 3A (4)2 + 2B (4) + C at time t = 4 sec. V = 48 A + 8B + C Considering A = 1, B = 4, C = –2 V = 48 (0) + 8 (4) + (–2) = 78 m/s Motion in a Straight Line Ex.13 In a car race, car A takes a time of t sec. less than car B at the finish and passes the finishing point with a velocity  m/s more than the car B. Assuming that the cars start from rest and travel with constant accelerations a1 and a2 respectively, show that v =  a1a2  t   Sol. Let the time taken by two cars to complete the journey be t1 and t2 and their velocities at the finish be 1 and 2 respectively.. Given that, t1  t 2  t and 1  2   .......... (1) Now 11 2 S1  s S1  a1t1 22 and (At start, v1 = v2 = 0) 1 2 Ss22  Ss  a2 t 2 2 Also, v 1  a1t1 and v 2  a 2 t 2 or v1t1 = a1t12 = 2s and v2t2 = a2t22 = 2s 2s 2s  t1  v and t 2  v 1 2  1 1 So, t 2  t1  2s  v  v  1  2 From equations (1) and (4) we have  1 1 2s   t  v 2 v1   v  ...... (2) or 2s  t  v 1v 2   v 1v 2     t1t 2  ............ (3) Hence a1t12  a2 t 22  2s ............ (4)  v  v2  2s  1 t  v 1v 2  or v v  or v   1 2  t   2s   v 12 v 22   t  (2s)2  × t = (a1a 2 )  t ACCELERATION It is defined as the rate of change of velocity. (i) It is a vector quantity. (ii) Its direction is same as that of change in velocity and not of the velcoity (That is why, acceleration in circular motion is towards the centre). (iii) There are three ways possible in which change in velocity may occur. Types of acceleration (a) Instantaneous acceleration : It is defined as the acceleration of a body at some particular instant.    v dv Instantaneous acceleration = lim  t 0 t dt  (b) Average acceleration = aav      v v 2  v1  t t 2  t1 (c) Uniform acceleration : A body is said to have uniform acceleration if magnitude and direction of the acceleration remains constant during particle motion. Note : If a particle is moving with uniform acceleration, this does not necessarily imply that particle is moving in straight line. Motion in a Straight Line Example : Parabolic motion (d) Non-uniform acceleration : A body is said to have non-uniform acceleration, if magnitude or direction or both, change during motion. Note : (i) Acceleration is a vector with dimensions [LT–2] and SI units (m/s2) (ii) If acceleration is zero, velocity will be constant and motion will be uniform. (iii) However if acceleration is constant acceleration is uniform but motion is non-uniform and if acceleration is not constt. both motion and acceleration are non-uniform.   (viii) The slope of a  t curve, i.e. d a / dt is a measure of rate of non-uniformity of acceleration. However we do not define this physical quantitiy as it in not involved in basic laws or equation of motion (ix) Acceleration can be positive or negative. Positive acceleration means velocity is increasing with time while negative acceleration called retardation means velocity is decreasing with time. Special note - I  (iv) If a force F acts on a particle of mass m then   by Newton’s law a  F / m   (v) As by definition, v  ds so, dt    dv dv  dt  dt ( 0) i.e. if s is given as a function of time, second time derivative of displacement gives acceleration. (vi) If velocity is given as function of position then dx [as dt Howeverm, if motion is uniform translatory, both these will still be equal but zero. Special note - II dv dv dx .  dt dx dt  (i) d| v |  0 while =v]        dv d  ds  d2 s  a   dt dt  dt  dt 2   dv av dx  (without change in direction) v  | v | n̂  d| v | dv d  [as n is constt.]   [| v | n̂]  n̂ dt dt dt  by the chain rule a  If the motion of a particle is accelerated translatory dt  (vii) As acceleration a  d v , the slope of velocity dt   time graph gives acceleration i.e. a  d v = tan dt  (ii) d| v |  0 while dt  dv 0 dt (it is possible.)  dv 0 dt (it is not possible.) EFFECTIVE USE OF MATHEMATICAL TOOLS IN SOLVING PROBLEMS OF ONE-DIMENSIONAL MOTION If displacement-time equation is given, we can get velocity-time equation with the help of differentiation. Again, we can get acceleration-time equation with the help of differentiation. If acceleration-time equation is given, we can get velocity-time equation by integration. From velocity equation, we can get displacement-time equation by integration. Motion in a Straight Line dx  x Sol. (i) Given that v   x or dt Ex.14 The displacement of a particle is given by y = a + bt + ct2 + dt4. Find the acceleration of a particle. dx  x Hence   dt x dx 0 x  or t    dt 0 2 x   t or x   2 t 2 / 4 dy d  a  bt  ct 2  dt 4  b  2ct  4dt 3 Sol. v  dt dt dv a  2c  12dt 2 dt Velocity Ex.15 If the displacement of a particle is (2t2 + t + 5) meter then, what will be acceleration at t = 5 sec. (ii) Time taken to cover first s metres dx d  2t 2  t  5  4t  1 m / s and Sol. v  dt dt dv d a  4t  1 a a=44 m/s m / 2s dt dt Ex.16 The velocity of a particle moving in the x direction varies as V   x where  is a constant. Assuming that at the moment t = 0 the particle was located at the point x = 0. Find the acceleration. dv d 1 dx  x   . x 1/ 2 . dt dt 2 dt 1 2 . . x a     = 2 2 x Sol. a  Ex.17 The velocity of any particle is related with its displacement As; x  v  1 , Calculate acceleration at x = 5 m. Sol. x  v  1 x2  v  1 v  x2  1 Therefore dv d 2 dx a  x  1  2x  2x v  2x x 2  1 dt dt dt at x = 5 m dx 1 2   t and dt 2 d2 x 1 2   Acceleration dt 2 2 s 2 t 2 4 v total dis tan ce total time positive constant. Assuming that at the moment t = 0, the particle was located at x = 0 find, (i) the time dependance of the velocity and the acceleration of the particle and (ii) the mean velocity of the particle averaged over the time that the particle takes to cover first s metres of the path. = 4s or 2 s t or v  2 s 2 s  1 s  2 Ex.19 A particle moves in the plane xy with constant acceleration a directed along the negative y-axis. The equation of motion of the particle has the form y = px – qx2 where p and q are positive constants. Find the velocity of the partcle at the origin of coordinates. Sol. Given that y = px – qx 2 and d2 y d2 x d2 x  dx   p  2qx  2q   2 2 2 dt dt dt  dt  2 2 or   dx  a  2q    2q2x  dt  2 dx 0 dt 2 (no acceleration along x-axis) and  a  2  5 25  1  240 m / s2 Ex.18 The velocity of a particle moving in the positive direction of x-axis varies as v=  x where  is t2  or Further,  Now 2x  a 2q or dx  dy   dt   p dt   x 0 x  or d2 y dt 2  a a 2q y  p x  a  y  p    2q    a ap 2  2x  2y      2q 2q   a p2  1    or  2q    Ans. Ex.20 A particle is moving in a plane with velocity given by u  u0i  a cos t j , where i and j are unit vectors along x and y axes respectively. If particle is at the origin at t = 0. Motion in a Straight Line Displacement in vertical direction y   a  cos t dt  a sin t .... (4) Eliminating t from equations (3) and (4) we get .... (5) (a) Calculate the trajectory of the particle. y  a sin  x / u0 (b) Find the distance from the origin at time 3 / 2 . Equation (5) gives the trajectory of the particle. (b) At time, Sol. (a) Given that u  u0i  a cos t j , t  3  / 2 Hence, velocity along x-axis ux  u0 .... (1) x  u0 3 / 2 and y  a sin3 / 2  a Velocity along y-axis uy  a cos t .... (2)  Distance of the particle from the origin We know that  ds or dt s    .dt So from equations (1) and (2), we have = 2 x y 2 =  3 u0  2   a2     2   Ans. Displacement at time t in horizontal direction x   u0 dt  u0 .t .... (3) SOME IMPORTANT GRAPHS RELATED TO MOTION All the following graphs are drawn for one-dimensional motion with uniform velocity or with constant acceleration. Motion in a Straight Line 3. Velocity-position relation MOTION WITH UNIFORM Motion under uniform acceleration is described by 1 2 at2v2 = u2 + 2as 1. The velocity time relation dv a dt dv = adt t2 is v2 v2 v1 t2 v dv   a dt t1 v 2  v 1  a t 2  t1 v2 v1 a t t2  x2 x1 or t2 dx   v dt  x 2  x1  ut  v1 x1  v2     2  v1 v 2  u2  2as v2 = u2 + 2as 4. Distance travelled in nth second of uniformly accelerated motion : a 2n  1 2 The calculation of speed and distance by acceleration-time graph: Let a particle be moving with uniform acceleration according to following a–t graph – v=u+at dv = a dt or dx = vdt v t  a x a dx   v dv x2 t1 2. Position-time relation dx dt a dx = v dv v 2  v1  a t 2  t1 If t1  0, v1  u and t 2  t, v 2  v By v  x1 v2 Snth  u  If the velocity of particle at time t1 is v1 and at time   x2 2a x 2  x1  v 22  v12 the following equations. s = ut + or v2 EQUATIONS OF MOTION : v = u + at dv dv dx dv  . v dt dt dx dx a ACCELERATION t1 1 2 at 2 1  x  x x2  ut  at2  1 2 0  s  ut  1 2 at 2 v u a t t2 t1  v u t2 dv   a dt t1 v  u  a t 2  t1 Therefore difference in magnitude of velocity (v – u) = AB x AD v – u = Area of rectangle ABCD = area under a – t graph Motion in a Straight Line Ex.21 The velocity acquired by a body moving with uniform acceleration is 20 m/s in first 2 sec and 40 m/s in first 4 sec. Calculate initial velocity. 1 2 2 Sol. (a) sT  at v v 2 1 Sol. a  t  t 2 1 a Now, v  u  at   40  20 20   10m / s2 42 2 v1  u  at1 20  u  10  2 20  u  20  u  0m/ s Ans Ex.22 A particle starts with initial velocity 2.5 m/s along the x direction and accelerates uniformly at the rate 50 cm/s2. Find time taken to increase the velocity to 7.5 m/s. 1 1.5 t 2 .... (1) 2 sT  Distance covered by car when car one overtakes the truck 1 2 t 2 .... (2) 2 sT  150 2  Divide equation (2) by equation (1) sT 1.5 150 20 4          1  sT 15 3 sc  1 2 t2 2  sT  150  150 4 1  1 sT 3 3 or sT  450 Distance travelled by car = 450 + 150 = 600 metre 1 2 2 (b) Now by equation (1) sT  at 1  1.5  t 2 2 450  2 2 t   1.5 450  Sol. v  u  at 7.5  2.5  0.5 t 50 t  10 sec 5 5.0  0.5 t Ex.23 A particle starts with a constant acceleration. At a time t second speed is found to be 100 m/ s and one second later speed becomes 150 m/s. Find acceleration of the particle. Sol. From equation (1) of motion v = u + at  100 = 0 + at 100 = at ... (1) t  300  2  24.5 sec Therefore car will overtake the truck after 24.5 second. Ex.25 A body travels a distance of 2 m in 2 seconds and 2.2 m in next 4 seconds. What will be the velocity of the body at the end of 7th second from the start. Sol. Here, case (i) s = 2m, t = 2s Now consider velocity one second later - case (ii) t = 2 + 4 = 6s v'  0  a t  1 Let u and a be the initial velocity and uniform acceleration of the body.  150  a t  1 ... (2) On subtracting equation (1) from equation (2) a  50 m / s 2 Ex.24 A truck starts from rest with an acceleration of 1.5 ms-2 while a car 150 metre behind starts from rest with an acceleration of 2 ms-2. (a) How long will it take before both the truck and car are side by side and (b) How much distance is travelled by each. we know that, s  ut  1 2 at 2 1 2 2 Case (i) 2  u  2  a  2 or 1 = u + a .... (1) 1 2 2 Case (ii) 4.2  u  6  a  6 or 0.7  u  3a ... (2) Motion in a Straight Line Ex.28 A bullet going with speed 350 m/s enters in a concrete wall and penetrates a distance of 5 cm before coming to rest. Find deceleration. Subtracting (2) from (1), we get 0.3  0  2a  2a or a  0.3 / 2  0.15ms2 Sol. Here, u = 350 m/s, From (i), u  1  a  1  0.15 u  1.15ms 1 v = 0 m/s and s = 5 cm, a=? For the velocity of body at the end of 7th second, we have u  1.15ms1;a  0.15ms2, v = ?, t = 7s As, v = u + at v = 0.1 m/s  v = 1.15 + (–0.15) x 7 Ex.26 A body travels a distance of 20 m in the 7th second and 24 m in 9th second. How much distance shall it travel in the 15th second? Sol. Here, s7 = 20 m ; s9 = 24 m, s15 = ? Let u and a be the initial velocity and uniform acceleration of the body. We know that, sn  u   s7  u  a 2n  1 2 a 13a 2  7  1 or 20  u  .... (i) 2 2 a 2 and s9  u  (2  9  1) or24 = u + Subtracting (ii) form (i), we get or a = 2 ms–2 Putting this value in (i), we get 20  u  13 2 2 a ... (ii) 4 = 2a a u 2 u2 2s = 12.25 × 105 m/sec2 Ex.29 A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of 54 km/h and the brakes cause a deceleration of 6.0 m/s2, find the distance travelled by the car after he seeds the need to put the brakes on or 5 m/ s 18 (2 × 15 – 1) = 7 + s1 = 15 x 0.2 = 3.0 meter × 29 5 m/s 18 v=0 Using u = 15 m/s, 2 a = 6 m/s2 s2 = ? 2 v = u - 2as 0 = (15)2 – 2 x 6 x s2 12 s2 = –225  s2  225  18.75metre 12 Distance travelled by the car after driver sees the need for it s = s1 + s2 s = 3 + 18.75 = 21.75 metre. Ans. = 15 m/s After applying the brakes; 2 2 Deceleration; a = 6 m/s2 v = 0 s = ? Using v2 = u2 - 2as 0 = (12)2 – 2 x 6 s 144 = 2 x 6s 144  12m 12 350  350 2  0.05 s1 = ut s15 = 36 m Ans. Ex.27 A person travelling at 43.2 km/h applies the brakes giving a deceleration of 6 m/s2 to his scooter. How far will it travel before stopping? s a= a u = 54 km/h = 54  –1 Sol. Here, u  43.2 km / h = 43.2  u  2as or 2 0  u2  2as Sol. Distance covered by the car during the application of brakes by driver - or 20 = u + 13 or u = 20 – 13 = 7 ms Hence, s15 = 7 2 By using v 2  u2  2as Ans. MOTION UNDER GRAVITY The most important example of motion in a straight line with constant acceleration is motion under gravity. In case of motion under gravity. (1) The acceleration is constant, i.e., a = g = 9.8 m/s2 and directed vertically downwards. (2) The motion is in vacuum, i.e., viscous force or thrust of the medium has no effect on the motion. [1] Body falling freely under gravity : Taking initial position as origin and downward direction of motion as positive, we have u=0 a = +g Motion in a Straight Line [as body starts from rest] [as acceleration is in the direction of motion] So, if the body acquires velocity v after falling a distance h in time t, equations of motion, viz. 1 2 v = u + at; s  ut  at 2 and reduces to v = gt ... (1) h v 2  u2  2as 1 2 gt ..... 2 (2) and v 2  2gh ..... (3) These equations can be used to solve most of the problems of freely falling bodies as if. (i) If the body is dropped from a height H, as in time t is has fallen a distance h from its initial position, the height of the body from the ground will be  1 h'  H  h with h    gt 2 2  1 2 (ii) As h   2  gt , i.e., h  t 2 , distance fallen in time,   t,2t,3t etc., will be in the ratio of 1 2 : 22 : 32, i.e., square of integers. (iii) The distance fallen in the nth sec, h n  h n 1 1  g n 2 2 1 2  g n 1 2 1  g 2n  1 2 So distances fallen in 1st, 2nd, 3rd sec etc. will be in the ratio of 1 : 3 : 5 i.e., odd integers only. [2] Body projected vertically up : Taking initial position as origin and direction of motion (i.e., vertically up) as positive. here we have v=0 [at highest point velocity = 0] a = – g [as acceleration is downwards while motion upwards] If the body is projected with velocity u and reaches the highest point at a distance h above the ground in time t, the equations of motion viz., v = u + at; s  ut  reduces to 0 = u – gt 1 2 at 2 and v 2  u2  2as 1 h  ut  gt 2 2 Motion in a Straight Line Substituting the value of u from first equation in second and rearranging these, u = gt ... (1) 1 2 gt 2 ... (2) and u2  2gh ... (3) h 2 and 0 = u – 2gh These equations can be used to solve most of the problems of bodies projected vertically up as if. IMPORTANT POINTS (1)In case of motion under gravity for a given body, mass, acceleration, and mechanical energy remain constant while speed, velocity, momentum, kinetic energy and potential energy change. (2)The motion is independent of the mass of the body, as in any equation of motion, mass is not involved. This is why a heavy and lighter body when released from the same height, reach the ground simultaneously and with same velocity. i.e t  2h / g and v  2gh However, momentum, kinetic energy or potential energy depend on the mass of the body (all  mass) (3)As from equation (2) time taken to reach a height h, tu  2h / g Similarly, time taken to fall down through a distance t u  tD  2h / g h, tD  2h / g so So in case of motion under gravity time taken to go up a height h is equal to the time taken to fall down through the same height h. (4)If a body is projectd vertically up and it reaches a height h, then u  2gh and if a body falls freely through a height h, then v 2gh  u So in case of motion under gravity, the speed with which a body is projected up is equal to the speed with which it comes back to the point of projection. Motion in a Straight Line Ex.32 A ball is thrown vertically upwards from a bridge with an initial velocity of 4.9 m/s. It strikes the water Ex.30 A juggler throws balls into air. He throws one after 2s. If acceleration due to gravity is 9.8 m/s 2 whenever the previous one is at its highest point. (a) What is the height of the bridge? (b) With which How high do the balls rise if he throws n balls each velocity does the ball strike the water? sec. Acceleration due to gravity is g. Sol. Taking the point of projection as origin and Sol. Since the juggler is throwing n balls each second downward direction as positive, and he throws second ball when the first ball is at  1 2 the highest point, so time taken by each ball to reach (a) Using s  ut   2  at we have   the highest point is t = 1/n  1 h  4.9  2    9.8  22  9.8m 2 Taking vertical upward motion of ball up to the highest point, we have (u is taken to be negative as it is upwards) u = 0, a = – g, t = 1/n, u = ? (b) Using v = u + at As v = u + at or u = g/n so Also 0 = 0 u + (–g) 1/n 2 2 v = u + 2as, 2 so 0 = u – 2gh i.e., h = (u2/2g) = g/(2n2) Ex.31 A ball is projected vertically up with an initial speed of 20 m/s on a planet where acceleration due to gravity is 10 m/s2. v = –4.9 + 9.8 x 2 = 14.7 m/s Ex.33 A rocket is fired vertically up from the ground with a resultant vertical acceleration of 10 m/s2. The fuel is finished in 1minute and it continues to move up. (a) What is the maximum height reached? (b) After how much time from then will the maximum height be reached? (Take g = 10 m/s2) (a) How long does it takes to reach the highest point? Sol. (a) The distance travelled by the rocket during burning interval (1 minute = 60 s) in which resultant (b) How high does it rise above the point of acceleration is vertically upwards and 10 m/s2 will be projection? (c) How long will it take for the ball to reach a point 10 m above the point of projection? Sol. As here motion is vertically upwards, a = –g and v = 0 (a) From 1st equation of motion, i.e., v = u + at, 0 = 20 – 10t i.e. t = 2 sec. Ans. h1  0  60  1/ 2  10  602  18000m Velocity acquired by it is v = 0 + 10 x 60 = 600 m/s ... (2) After one minute the rocket moves vertically up with initial velocity of 600 m/s and continues till height h2 till its velocity becomes zero. 0  600 (b) Using v2 = u2 + 2as ... (1) 2  2gh 2 [as g = 10 m/s2] 0 = (20)2 – 2 x 10 x h or h2  18000m i.e h = 20 m. Ans. From equations (1) and (3) the maximum height reached by the rocket from the ground is 1 2 1 2 (c) Using s  ut  at 2 , 10 = 20t (– ) x 10 x t2 2 i.e. t – 4t + 2 = 0 or t  2 2 , i.e. t = 0.59 sec. or 3.41 sec. i.e., there are two times, at which the ball passes through h = 10 m, once while going up and then coming down. ... (3) H  h1  h2  18  18  36km (b) The time to reach maximum height after burning of fuel is 0 = 600 – gt t = 60 s After finishing fuel the rocket goes up for 60 s. Ex.34 A body is released from a height and falls freely towards the earth. Exactly 1 sec later another body is released. What is the distance between the two bodies after 2 sec the release of the second body, if g = 9.8 m/s2. Sol. The 2nd body falls for 2s, so 1 h2  g 2 2 2 ... 2 ... 1 g 3 2  22 2 (2) [ u = 0 as the body starts from rest] ... (1) Now, as the distance covered in (t – 1) second is 2 ... (2) So from Equations (1) and (2) distance travelled in the last second. 1 1 2 '  gt 2  g t  1 hh – hh’ 2 2 1 i.e., h –h h’h  g 2t  1 2   But according to given problem as (h –h’) 11  1     1   2 or  2  gt  g 2t  1   1 [as from equation (1) h    gt 2 ] 2 or t 2  4t  2  0 Effect of Thrust : in a medium of density  , thrust will be Vg and will oppose the weight. So there are 3 possibilities: (a)If    , thrust will be greater than weight and so there will be a net force (Th–W) opposite to gravity and so the body will experience an acceleration. or t = 0.59 s h 2 Th  W m opposite to gravity Th V g  – gg  g   1 = m  g as W  mg  V   g’    = g    1   (as m = V) ...... (1) So, if a body is at rest it will be opposite to acceleration due to gravity (as in the case of a hydrogen filled balloon) and if the body falls through a liquid with an initial velocity the motion will be retarded [with retardation given by equation (1)], so body will stop after travelling distance and finally will move up [with acceleration given by equation (1)]. (b)If    the net force F  W  Th  Vg  Vg  0 or mg’ = 0 , or t  [4  ( 4 2  4  2)] / 2 or t  2  2 Medium effects the motion of a body falling freely under gravity due to thrust and viscous drag. g’     i.e.,  2 h    g 2t  1 2 2  × (9.8) × (3.41)2 = 57 m If a body of volume V and density  is dropped  4.9  5  24.5m 1 2 gt 2 1 2 2 MOTION UNDER GRAVITY Sol. If the body falls a height h in time t, then h  g t 1 h’ 1 EFFECT OF MEDIUM ON Ex.35 If a body travels half its total path in the last second of its fall from rest, find : (a) The time and (b) height of its fall. Explain the physically unacceptable solution of the quadratic time equation. (g = 9.8 m/s2) h t = 3.41 s (1)  Separation between two bodies after 2 sec the d  h1  h2 release of 2nd body,  so and h = While 1st has fallen for 2 + 1 = 3 sec so 1 h1  g 3 2 Motion in a Straight Line 0.59 s is physically unacceptable as it gives the total time t taken by the body to reach ground lesser than one sec while according to the given problem time of motion must be greater than 1s. or 3.41 s i.e. g ‘ = 0 ... (2) So, the body will be in natural equilibrium i.e., it will remain at rest or in uniform motion as the case may be.    , then as thrust will be lesser than weight, (c) If net force will be downwards and will be F = mg’ = mg – Th or   Th g’  g 1   g  g  m  .... (3)       (i) There is nothing in absolute rest or absolute motion. (ii) Motion is a combined property of the object under study and the observer. i.e., now body will be accelerated down with  Motion in a Straight Line RELATIVE - VELOCITY g  g 1     , which is less than g i.e. acceleration g’  the effect of thrust here is to decrease acceleration due to gravity. Effect of Viscous Force : When a spherical body of radius r falls wtih a velocity in a medium of viscosity  , the viscous force opposing the motion, according to Stokes-law is given by 6rv. This force is velocity dependent and non conservative. So when a spherical body Example (i) A book placed on the table in a room is at rest, if it is viewed from the room but it is in motion, if it is viewed from the moon (another frame of reference). The moon is moving w.r.t. the book and the book w.r.t. the moon. (ii) A robber enters a train moving at great speed with respect to the ground, brings out his pistol and says “Don’t move, stand still”. falls freely in a medium   p due to force mg the The passengers stand still. The passengers are at rest with respect to the robber but are moving with respect to the rail track. velocity will increase and as velocity increases the (iii) Relative motion means, the motion of a body viscous force  v will also increase. After some with respect to another. Now if VA and VB are   velocities of two bodies relative to earth, the velocity time a stage will be reached when viscous force    of B relative to A will be given by VBA  VB  VA exactly balances the net downward force mg. Then body will fall with a constant velocity called terminal Note : velocity. Thus if v T is the terminal velocity.. (a)If two bodies are moving along the same line in 6r vT = mg , i.e. v T  mg / 6r same direction with velocities VA and VB relative But for a spherical body as to earth, the velocity of B relative to A will be given  4 3  2  2 g by VBA  VB  VA . m    r  vT   r .... (4) 3 9  However if, along with viscous force thrust is also taken into account which is usually the case, by placing g’ If it is positive the direction of VBA is that of B and if negative the direction of VBA is opposite to that of B. (b)However, if the bodies are moving towards or away from each other, as direction of VA and VB are opposite, velocity of B relative to A will have magnitude VBA  VB   VA  VB  VA and directed towards A or away from A respectively. (c) In dealing the motion of two bodies relative to   g  g 1     we get,  each other Vrel is the difference of velocities of two in equation (4),  2      vT   r2  g 9    ... (5) bodies, if they are moving in same direction and is the sum of two velocities if they are moving in opposite direction. (d)If a man can swim reltive to water with velocity  V and water is flowing relative to ground with Motion in a Straight Line on a rough day, let v’ be the velocity of air current. As in going across the lake, the air current helps velocity VR , velocity of man relative to ground Vm the motion, so time taken is will be V  Vm  VR i.e. Vm  V  VR In coming back, as the air current opposes the So, if the swimming is in the direction of flow of water Vm  V  VR motion, so time taken is t 2  And if the swimming is opposite to the flow of water Total time in going and coming back on a rough day         Vm  V  VR  (e)If a boy is running with velocity Vrel. on a train  moving with velocity V relative to the ground. The  velocity of boy relative to ground, V will be given       by Vrel.  v  V  v  Vrel.  V So, if the boy is running on the train in the direction of motion of train v  Vrel.  V And if the boy is running on the train in a direction opposite to the motion of train v  Vrel.  V (f) Suppose a train having length  is crossing a bridge of length L with constant speed v. Then the time taken by the train to cross the bridge will be : t=  L v t R  t1  t 2  2v 2 2 (v  v' )  t1   v  v'  v  v' 2 v[1  (v' / v )2 ] .......(2) from (i) and (ii), we have tR 1  1 t Q [1  ( v ' / v' )2 ] or tR > tQ Therefore the time taken to complete the journey on quiet day is less than on a rough day. Crossing of flowing river: Let AB and CD be two banks of a river. The width of the river is PQ = h. It is flowing in the direction of arrow. v is velocity of water flowing in river and u is velocity of swimmer or boat with respect to still water. (a) If a boat rows with a speed u at an angle  with the direction of width of the river, then the time taken by the man to cross the river along PQ is Ex.36 A steam boat goes across a lake and comes back : (a) on a quiet day when the water is still and (b) on a rough day when there is a uniform current so as to help the journey onward and to impede the journey backward. If the speed of launch on both days was same, in which case will it complete the journey in lesser time? t h  ucos  h 2 u  v2 where sin   (b) If a boat moves with a velocity u along the direcion of the width, then the resultant velocity of the boat crossing the river makes an angle  with PQ. Sol. Let be the width of lake and v be the velocity of steam boat. On a quiet day, time taken by steam boat in going and coming back is. tQ =     2 v v v ......(1) v u  Resultant velocity  u2  v 2 Motion in a Straight Line (c)  Distance traversed by the man h  t u2  v 2 cos  (d) Time taken in crossing the river t h 2 cos  u  v 2  h u This will be minimum time for crossing the river. (e) Distance traversed in the direction of flow of h v  v  h  river  u u (f) Thus to cross the river in minimum time the boat is rowed in a direction perpendicular to the velocity of water. To cross the river from one point P to an opposite point O on the other bank the direction of rowing makes an angle  with PQ such that v   sin1   u (g) To across the river from one point on one bank to an opposite point on the other bank the velocity   of boat u makes an angle  2      Ex.37 A swimmer can swim in still water at a rate 4 km/hour. If he swims in a river flowing at 3 km/ h and keeps his direction (with respect to water.) perpendicular to the current. Find his velocity with respect to the ground. Sol. The velocity of the swimmer with respect to water  is v SR  4.0km / hr in the direction perpendicular to the river. The velocity of river with respect to the  ground is v RG  3.0km / hr along the length of river.. The velocity of the swimmer with respect to the  ground is v SG where    VSG  VSR  VRS 2 2 VSG  VSR  VRG  42  3 2  16  9  25  5km / hr The angle  made with the direction of flow is To cross the river in minimum time   0 and h t u Rain problems  If rain is falling vertically with a velocity VR and an  observer is moving horizontally with speed Vm , the velocity of rain relative to observer :  VSR  1 4 = tan–1  V   tan  3     RG  Ex.38 A man can swim in still water at a speed of 3 km/hr. He wants to, cross a 500 m wide river flowing at 2 km/hour. He keeps himself always at an angle of 1200 with the river flow while swimming. (a) Find the time he takes to cross the river. (b) At what point on the opposite bank will he arrive.  Sol. Width of river AB = d = 500 m  1 2 km.    vRM  VR  Vm which by law of vector addition has magnitude  VRM  VR2  VM2 V  1 M and direction   tan  V  with vertical.  R  Vm = 3 km/hr. Velocity of man in still water Vr = 2 km/hr. Velocity of river V = resultant velocity of man in flowing river.  V  Vx ˆi  Vy ˆj Now, Vx  Vr  Vm sin300 = 1 1 2  3   km / hr 2 2 Vy  Vm cos300  3 3 km / hr 2 Displacement along Y-axis, d = Vy x t or Motion in a Straight Line Ex.40 A girl standing on a road has to hold her umbrella at 300 with the vertical to keep the rain away. She throws the umbrella and starts running at 10 km/ hr. She finds that raindrops are hitting her head vertically. Find the speed of raindrops with respect to (a) the road (b) the moving girl. Sol. Suppose, the velocity of rain with respect to girl 1 d t  2 Vy 3 3 2  t 1 3 3 hr = VRG The velocity of rain with respect to the ground Displacement along X-axis, BC  Vx  t = VRG 1 1 1   km = 2 3 3 6 3 The velocity of girl with respect to ground Ex.39 A man is walking on a level road at a speed of 3 km/hr. Raindrops fall vertically with a speed of 4 km/hr. Find the velocity of raindrops with respect to the men. Sol. If we consider velocity of rain with respect to the man is v km/h.  VGg  10 km / hr.    v RG  v R  v G    v RG  v R  v g ... ... (1) (2)    v Gg  v G  v g ... (3) Relative velocity of man w.r.t. ground    v mg  v m  v g ......... (1)    Velocity of rain w.r.t. ground v rg  vr  v g .... (2) Velocity of rain w.r.t. man    v rm  v r  v m On subtracting equation (1) from equation (2)    v rm  v rg  v mg On adding equation (1) and equation (3)      v RG  v Gg  v R  v g  v Rg (a) By triangle AOB sin30 0  2 v rm  v rg2  v mg  42  3 2  5km / hr. Direction : 3 tan   4 or 3   tan1   4 VRg  AB 10  OB VRg 10 10   20 km / hr sin30 0 1/ 2 (b) Now, taking VRG  cot 30 0 VGg VRG  3 10 or VRG  10 3 Km / hr