Kinematics Notes

KINEMATICS Page # 1 KINEMATICS 1. REST AND MOTION : * An object is said to be in motion wrt a frame of reference S1, when its location is changing with time in same frame of reference S1. * Rest and motion are relative terms. *Absolute rest and absolute motion have no meaning. Motion is broadly classified into 3 categories. 1. Rectilinear and translatory motion. 2. Circular and rotatory motion. 3. Oscillatory and vibratory motion. 1.1 Rectilinear or 1-D Motion When a particle is moving along a straight line, then its motion is a rectilinear motion. Parameters of rectilinear motion or translatory motion or plane motion : (A) Time : * It is a scalar quantity and its SI unit is second(s). * At a particular instant of time, a physical object can be present at one location only. * Time can never decrease. (B) Position or location - It is defined with respect to y A C some reference point (origin) of given frame of reference.  Consider a particle which moves from location r1 (at time t1) r1 B  to location r2 (at time t2) as shown in the figure below, following path ACB. (C) Distance : The length of the actual path traversed by the particle is termed as its distance. Distance = length of path ACB. * Its SI unit is metre and it is a scalar quantity. * It can never decrease with time. r2 x (D) Displacement : The change in position vector of the particle for a given time interval is known as its displacement.     AB  r  r2  r1 * Displacement is a vector quantity and its SI unit is metre. * It can decrease with time. For a moving particle in a given interval of time * Displacement can be +ve, –ve or 0, but distance would be always +ve. * Distance  Magnitude of displacement. * Distance is always equal to displacement only and only if particle is moving along a straight line without any change in direction. (E) Average speed and average velocity : Average speed and average velocity are always defined for a time interval. Total dis tan ce travelled s  Time int erval t     r r Displacement r Average velocity (v av )    2 1 Time int erval t t2  t1 Average speed(vav )  * Average speed is a scalar quantity, while average velocity is a vector quantity. Both have the same SI units, i.e., m/s. For a moving particle in a given interval of time * Average speed can be a many valued function but average velocity would be always a singlevalued function. * Average velocity can be positive, negative or 0 but average speed would be always positive. KINEMATICS Page # 2 (F) Instantaneous speed and instantaneous velocity Instantaneous speed is also defined exactly like average speed i.e. it is equal to the ratio of total distance and time interval, but with one qualification that time interval is extremely (infinitesimally) small. The instantaneous speed is the speed at a particular instant of time and may have entirly different value than that of average speed. Mathematically. s ds v  lim  ...(4) dt s0 t When s is the distance travelled in time t. As t tends to zero, the ratio defining speed becomes finite and equals to the first derivative of the distance. The speed at the moment 't' is is called the instantaneous S Distance B speed at time 't'. D S  A O C t t t  t time t Instantaneous speed is equal to the slope of the tangent at given instant. On the distance - time plot, the speed is equal to the slope of the tangent to the curve at the time instant 't'. Let A and B point on the plot corresponds to the time t and t + t during the motion. As t approaches zero, the chord AB becomes the tangent AC at A. The slope of the tangent equal ds/dt, which is equal to the intantaneous speed at 't'. v = tan = DC ds = AC dt (G) Instantaneous velocity : Instantaneous velocity is defined exactly like speed. It is equal to the ratio of total displacement and time interval, but with one qualification that time interval is extremely (infinitesimally) small. Thus, instantaneous velocity can be termed as the average velocity at a particular instant of time when  t tend to zero and may have entirely different value that of average velocity : Mathematically. The magnitude of average velocity |vavg| and average speed vavg may not be equal, but magnitude of instantaneous velocity |v| is always equal to instantaneous speed v. dx x 2 x  dt  dx x 1 2 t   0 dt x  1  –   t  x 1  – S Position/displacement r dr v  lim  dt t 0 t As  t tends to zero, the ratio defining velocity becomes finite and equals to the first derivative of the position vector. The velocity at the moment 't' is called the instantaneous velocity or simply velocity at time 't'. B B' D S  A O t t C t  t time t Instantaneous velocity is equal to the slope of the tangent at given instant. 1 1 +1=tx= 1– t x Average and instantaneous acceleration. When the velocity of a moving object/particle changes with time, we can say that it is accelerated. Average acceleration, KINEMATICS Page # 3     v  v1 v Change in velocity (aav )  2   t2  t1 t Time int erval Instantaneous acceleration,    dv (a)  lim aav  = Rate of change of velocity t  0 dt Acceleration is a vector quantity whose direction is same as that of change in velocity vector. Its SI unit is m/s2. * When direction of acceleration and velocity are opposite to each other, then acceleration is termed as retardation.     dv  dv d2r a    v  * dt dr d t2 Deduce the following equations for unifromly accelerated motion by using intergration technique. 1 2 at (C) v2 – u2 = 2as 2 First equation of motion. Acceleration is defined as (A) v = u + at (B) s = ut + (D) snth = u + a (2n – 1) 2 dv dt or dv = adt ...(1) When time = 0, velocity = u (say) When time = t, velocity = v (say) Integrating equation (1) within the above limits of time and velocity, we get a v  t  dv  a dt or u t [ v]uv  a 0  dt  at t 0 0 or v – u = a(t – 0) or v = u + at Second equation of motion. Velocity is defined as ...(2) ds dt or ds = v dt = (u + at) dt ...(iii) When time = 0, displacement travelled = 0 When time = t, displacement travelled = s (say). Integrating equation (3) within the above limits of time and distance, we get v s t t t  ds  (u  at) dt  u  dt  a  t dt 0 or 0 0 0 or [s]0s  u[t]0t  t2   a   2  t 0  t2  s – 0 = u (t – 0) + a  2  0   1 2 at ...(4) 2 Third equation of motion. By the definitions of acceleration and velocity, dv dv ds dv a    v dt ds dt ds or ads = vdv ...(5) When time = 0, velocity = u, displacement travelled = 0 When time = t, velocity = v, displacement travelled = s (say) Integrating equation (5) within the above limits of velocity and displacement, we get or s = ut + KINEMATICS s Page # 4 v  a ds   v dv 0 s  a ds  or u v v 0  v dv a[s]0s or u v2 u2  or 2as = v2 – u2 2 2 or v2 – u2 = 2as Fourth equation of motion. By definition of velocity, or  v2     2 u a[s  0]  v ...(6) ds dt or ds = vdt = (u + at) dt ...(7) When time = (n – 1) second, displacement travelled = sn – 1 (say). When time = n second, displacement travelled = sn (say) Integrating equation (7) within the above limits of time and distance, we get sn  sn 1 n ds   (u  at)dt n 1 = u[n –(n – 1)] + snth  u  where 2. or [s]ssn n 1 n u a [n2 – (n – 1)2] 2 a (2n  1) 2  n 1 n n dt  a  t dt or sn  sn 1  u[t]nn 1 n 1 =u+  t2   a   2 n 1 a 2 [n – (n2 – 2n + 1)] 2 ...(8) snth = sn – sn –1 = displacement in nth second. MOTION UNDER GRAVITY : B I FORMAT : (When a body is thrown vertically upward) It includes two types of motion upward (i) Deaccelerated motion from A to B because the direction motion downward motion (accelerated motion) of velocity and acceleration is opposite. So speed (deaccelerated decreases motion) u (ii) Accelerated motion from B to C because the direction of C velocity and acceleration is same (downward). So speed increases A (a) Time of flight : It is the time taken by the particle to reach the ground. If the particle is thrown vertically upward with initial velocity u then ui = u a = – g (take downward direction negative) from equation S = ut + 1 2 at  Snet = 0 (when particle again reaches the ground) 2 t = T (time of flight) 0 = uT – 2u 1 2 gT  T = g 2 (b) Maximum Height : from v2 u2 + 2as at maximum height v = 0, s = Hmax  0 = u2 –2 gHmax Hmax = u2 2g KINEMATICS Page # 5 (c) Final velocity from v = u + at v = vf a=–g t=T= 2u  g  2u  vf = u – g    g vf = – u i.e. the body reaches the ground with the same speed with which it was thrown vertically upwards as it thrown vertically upward. II Format (Free fall) : A body released near the surface of the earth is accelerated downward under the influence of force of gravity. (0,0) u=0 (a) Time of Flight : from equation S = ut + 1 2 at 2 H S = – H, u = 0, a = – g t = T (Let assume)  – H = (0)T – 1 2 gt  T = 2 2H g (b) Final Velocity when body reaches the ground from v2 – u2 = 2as s=–H v = vf u = 0 a = – g  v f 2 – 0 = 2 (–g) (–H)  vf = 2gH TWO DIMENSIONAL MOTION OR MOTION IN A PLANE # VECTOR # 1. SCALAR : In physics we deal with two type of physical quantity one is scalar and other is vector. Each scalar quantity has a magnitude and a unit. For example mass = 4kg Magnitude of mass =4 and unit of mass = kg Example of scalar quantities : mass, speed, distance etc. Scalar quantities can be added, subtracted and multiplied by simple laws of algebra. 2. VECTOR : Vector are the physical quantites having magnitude as well as specified direction. For example : Speed = 4 m/s (is a scalar) Velocity = 4 m/s toward north (is a vector) If someone wants to reach some location then it is not sufficient to provide information about the distance of that location it is also essential to tell him about the proper direction from the initial location to the destination.   The magnitude of a vector ( A ) is the absolute value of a vector and is indicated by | A | or A. Example of vector quantity : Displacement, velocity, acceleration, force etc. Knowledge of direction N W E S KINEMATICS Page # 6 Equality of Vectors. Vectors having equal magnitude and same direction are called equal vectors  C  A Collinear vectors :  B    if | A| |B| | C|   B  C  and A    then A  B  C Any two vectors are co-linear then one can be express in the term of other.   a = b (where  is a constant) Co-initial vector : If two or more vector start from same point then they called co-initial vector.  A A  B C  C e.g. O here A, B, C, D are co-initial. B  D D Coplanar vectors : Three (or more) vectors are called coplanar vectors if they lie in the same plane or are parallel to the same plane. Two (free) vectors are always coplanar. Important points  If the frame of reference is translated or rotated the vector does not change (though its components  B ve ct or may change).  A S' O' O S Two vectors are called equal if their magnitudes and directions are same, and they represent values of same physical quantity. Multiplication and division of a vector by a scalar    Multiplying a vector A with a positive number  gives a vector (B   A ) whose magnitude become    times but the direction is the same as that of A . Multiplying a vector A by a negative number  gives a    vector B whose direction is opposite to the direction of A and whose magnitude is – times | A | .   1 . The division of vector A by a non-zero scalar m is defined as multiplication of A by m   At here A and B are co-linear vector 3. LAWS OF ADDITION AND SUBTRACTION OF VECTORS : 3.1 Triangle rule of addition : Steps for additing two vector representing same physical quantity by triangle law. KINEMATICS Page # 7 (i) Keep vectors s.t. tail of one vector coincides with head of other. (ii) Join tail of first to head of the other by a line with arrow at head of the second. (iii) This new vector is the sum of two vectors. (also called reultant) D (i) Q. AD CD C A D (ii) AB B (iii) AB  CD  AD C B A Take example here. A boy moves 4 m south and then 5 m in direction 37° E of N. Find resultant displacement. 3.2 Polygon Law of addition :  P  a    b a  c  d b  c This law is used for adding more than two vectors. This is extension of triangle law of addition. We keep on arranging vectors s.t. tail of next vector lies on head of former. When we connect the tail of first vector to head of last we get resultant of all the vectors.  d d  c   c  b a  b  a  b          Note : P  (a  b)  c  d  (c  a  d)  d      a [Associative Law] 3.3 Parallelogram law of addition : Steps : (i) Keep two vectors such that there tails coincide. (ii) Draw parallel vectors to both of them considering both of them as sides of a parallelogram. (iii) Then the diagonal drawn from the point where tails coincide represents the sum of two vectors, with its tail at point of coincidence of the two vectors.  b  a (i)  b  a (ii)  b  a (iii)  b A D  a  a  b C   AC  a  b B         Note : AC  a  b and AC  b  a thus a  b  b  a [Cummutative Law] Note : Angle between 2 vectors is the angle between their positive directions.   Suppose angle between these two vectors is , and | a | a, | b | b (AD)2 Thus, = (AE)2 + (DE)2 = (AB + BE)2 + (DE)2 = (a + b cos )2 + (b sin )2 = a2 + b2 cos2 + 2ab cos  + b2 sin2 = a2 + b2 + 2ab cos  AD = a 2  b 2  2ab cos   b A  a  D  b a     B E a C KINEMATICS Page # 8   | a  b | a 2  b 2  2ab cos  b sin  DE angle  with vector a is tan  = = (a  b cos ) AE or Important points :  To a vector, only a vector of same type can be added that represents the same physical quantity and the resultant is also a vector of the same type.  As R = [A2 + B2 + 2AB cos]1/2 so R will be maximum when, cos  = max = 1, i.e.,  = 0º, i.e. vectors are like or parallel and Rmax = A + B.   | A || B | and angle between them  then R = 2 A cos  / 2   | A || B | and angle between them  – then R = 2 A sin  / 2      The resultant will be minimum if, cos  = min = – 1, i.e.,  = 180º, i.e. vectors are antiparallel and Rmin = A – B. If the vectors A and B are orthogonal, i.e.,  = 90º, R  A 2  B 2 As previously mentioned that the resultant of two vectors can have any value from (A – B) to (A + B) depending on the angle between them and the magnitude of resultant decreases as  increases 0º to 180º.   Minimum number of unequal coplanar vectors whose sum can be zero is three. 4. SUBTRACTION OF VECTOR : The resultant of three non-coplanar vectors can never be zero, or minimum number of non coplanar vectors whose sum can be zero is four.   Negative of a vector say – A is a vector of the same magnitude as vector A but pointing in a direction  opposite to that of A .         Thus, A – B can be written as A  (– B ) or A – B is really the vector addition of A and – B .  A  –A     Suppose angle between two vectors A and B is . Then angle between A and – B will be 180° –  as shown in figure.  180     B  A (a)    A  –B    S  A– B (b)  Magnitude of S  A – B will be thus given by   S = | A– B | = or S= A 2  B 2  2AB cos(180 – ) A 2  B 2 – 2 AB cos   For direction of S we will either calculate angle  or , where, ....(i) KINEMATICS or tan = B sin(180 – ) B sin  = A  B cos(180 – ) A – B cos  ...(ii) tan  = A sin(180 – ) A sin  = B  A cos(180 – ) B – A cos  ...(iii) UNIT VECTOR AND ZERO VECTOR  Unit vector is a vector which has a unit magnitude and points in a particular direction. Any vector ( A ) can be written as the product of unit vector (  ) in that direction and magnitude of the given vector..   A A  A  or   A A unit vector has no dimensions and unit. Unit vectors along the positive x-, y-and z-axes of a rectangular coordinate system are denoted by î, ĵ and k̂ respectively such that | î | = | ĵ | = | k̂ | = 1. y ĵ î x k̂ z A vector of zero magnitude is called a zero or a null vector. Its direction is arbitrary. Y u  O  u sin 5. Page # 9 + u cos  x Assume that effect of air friction and wind resistance are negligible and value of ‘acceleration due to  gravity g is constant. Take point of projection as origin and horizontal and vertical direction as +ve X and Y-axes, respectively. For X-axis For Y - axis ux = u cos, uy = u sin ax = 0, ay = – g, vx = u cos, and vy = u sin – gt, and 1 x = u cos × t y = u sint – gt2 2 It is clear from above equations that horizontal component of velocity of the particle remains constant while vertical component of velocity is first decreasing, gets zero at the highest point of trajectory and then increases in the opposite direction. At the highest point, speed of the particle is minimum. The time, which projectile takes to come back to same (initial) level is called the time of flight (T). At initial and final points, y = 0, 1 So u sint – gt2 = 0 2 2u sin 2u sin So, T= g g Range (R) The horizontal distance covered by the projectile during its motion is said to be range of the projectile  t = 0 and t  u 2 sin 2 g For a given projection speed, the range would be maximum for  = 45°. Maximum height attained by the projectile is R = u cos × T = KINEMATICS Page # 10 u 2 sin 2  2g at maximum height the vertical component of velocity is 0. u sin T Time of ascent = Time of descent = = g 2 H= Speed, kinetic energy, momentum of the particle initialy decreases in a projectile motion and attains a minimum value (not equal to zero) and then again increases.   is the angle between v and horizontal which decreases to zero. (at top most point) and again increases in the negative direction Projectile fired parallel to horizontal. As shown in shown figure suppose a body is projected horizontally with velocity u from a point O at a certain height h above the ground level. The body is under the influence of two simultaneous independent motions: u O (i) Uniform horizontal velocity u. x (ii) Vertically downward accelerated motion with y vx constant acceleration g. x P  h Under the combined effect of the above two motions, v the body moves along the path OPA. Trajectory of the projectile. After the time t, suppose the body reaches the point P(x, y). The horizontal distance covered by the body in time t is x x = ut  t= u The vertical distance travelled by the body in time t is given by 1 2 s = ut + at 2 1 2 1 2 gt or y = 0 × 1 + gt = 2 2 [For vertical motion, u = 0] 2 y or y = kx2 A Ground g = a constant] 2u2 As y is a quadratic function of x, so the trajectory of the projectile is a parabola. Time of flight. It is the total time for which the projectile remains in its flight (from 0 to A). Let T be its time of flight. For the vertical downward motion of the body, we use 1 2 s = ut + at 2 or h=0×T+ [Here k = R Y x   t  u    1 x  g  g     2  x2 2 u  2u  or vy 1 gT 2 2 or T= 2h g Horizontal range. It is the horizontal distance covered by the projectile during its time of flight. It is equal to OA = R. Thus R = Horizontal velocity × time of flight = u × T 2h g Velocity of the projectile at any instant. At the instant t (when the body is at point P), let the velocity of the projectile be v. The velocity v has two rectangular components: or R=u KINEMATICS Page # 11 Horizontal component of velocity, vx = u Vertical component of velocity, vy = 0 + gt = gt  The resultant velocity at point P is v v2x  v2y = u2  g2 t2 If the velocity v makes an angle  with the horizontal, then vy  gt  gt tan  = = or  = tan–1   vx u u 6. RELATIVE MOTION The word 'relative' is a very general term, which can be applied to physical, nonphysical, scalar or vector quantities. For example, my height is five feet and six inches while my wife's height is five feet and four inches. If I ask you how high I am relative to my wife, your answer will be two inches. What you did? You simply subtracted my wife's height from my height. The same concept is applied everywhere, whether it is a relative velocity, relative acceleration or anything else. So, from the above discussion we may now  conclude that relative velocity of A with respect of B (written as v AB ) is    v AB  v A – v B Similarly, relative acceleration of A with respect of B is    a AB  a A – a B If it is a one dimensional motion we can treat the vectors as scalars just by assigning the positive sign to one direction and negative to the other. So, in case of a one dimensional motion the above equations can be written as vAB = vA – vB and aAB = aA – aB Further, we can see that     v AB  – v BA or a BA  – a AB IMPORTANT NOTE : PROCEDURE TO SOLVE THE VECTOR EQUATION.    A  B  C ...(1) (a) Their are 6 variables in this equation which are following :  (1) Magnitude of A and its direction  (2) Magnitude of B and its direction  (3) Magnitude of C and its direction. (b) We can solve this equation if we know the value of 4 varibales [Note : two of them must be directions] (c) If we know the two direction of any two vectors then we will put them on the same side and other on the different side. For example    If we know the directions of A and B and C' s direction is unknown then we make equation as follows :    C  A –B (d) Then we make vector diagram according to the equation and resolve the vectors to know the unknown values.