Motion Notes

MOTION  Two dimensional motion A particle moving along a curved path in a plane has 2-dimensional motion. CONTENTS       Introduction  Circular Motion  Three dimensional motion Particle moving randomly in space has 3-dimensional motion. Types of motion Terms used to define motion  According to state of motion Uniform Motion Equations of Motion Body falling freely under gravity Various Motion Graphs Related to When a body does not change its position with time, we can say that the body is at rest, while if a body changes its position with time, it is said to be in motion.  An object is said to be a point object if it changes its position by distances which are much greater than its size.  A point or some stationary object with respect to which a body continuously changes its position in the state of motion is known as origin or reference point.  According to Directions  One dimensional motion is the motion of a particle moving along a straight line.  A body is said to be in a state of uniform motion if it travels equal distances in equal intervals of time.  If the time distance graph is a straight line the motion is said to be uniform motion. Non-uniform motion  A body has a non-uniform motion if it travels unequal distances in equal intervals of time. Ex. a freely falling body.  Time - distance graph for a body with nonuniform motion is a curved line. (i) Distance and displacement (ii) Speed and velocity (iii) Acceleration (i) Distance & Displacement  The path length between the initial and final positions of the particle gives the distance covered by the particle.  The minimum distance between the initial and final positions of a body during that time interval is called displacement  Distance and displacement both are measured in meter in m.k.s. system. Difference between displacement distance and  Distance travelled is a scalar quantity while displacement is a vector quantity. When a body continuously moves in the same Ex.2 A body is moving in a straight line. Its distances from origin are shown with time in Fig. A, B, C, D and E represent different parts of its motion. Find the following : straight line and in the same direction then displacement will be equal to the distance travelled. But if the body changes its direction (i) Displacement of the body in first 2 seconds. while moving, then the displacement is smaller than the distance travelled. (iii) Displacement in 7 seconds (ii) Total distance travelled in 7 seconds. Displacement  Distan ce Displacement in any interval of time may be zero, positive or negative where as distance cannot be negative.. Ex.1 A person travels a distance of 5 m towards east, then 4 m towards north and then 2 m towards west. (i) Calculate the total distance travelled. (ii) Calculate the resultant displacement. Sol. (i) Total distance travelled by the person = 5 m + 4 m + 2 m = 11 m (ii) To calculate the resultant displacement, we choose a convenient scale, where 1 cm represents 1 m. We draw a 5 cm long line AB towards east and then 4 cm long line BC towards north. Finally, a 2 cm long line CD towards west. The resultant displacement is calculated by joining the initial position A to the final position D. We measure AB = 5 cm. Since 1 cm = 1 m  5 cm = 5 m Hence, the displacement of the person Sol. (i) Displacement of the body in first 2s = 40m (ii) From t = 0 to t = 7 s, the body has moved a distance of 80 m from origin and it has again come back to origin. Therefore, the total distance covered = 80 × 2 = 160 m (iii) Since the body has come back to its initial position, the displacement is zero. (ii) Speed and Velocity  The 'distance' travelled by a body in unit time interval is called its speed. When the position of a body changes in particular direction, then speed is denoted by 'velocity'. i.e. the rate of change of displacement of a body is called its Velocity.   Speed is a scalar quantity while velocity is a vector quantity. = 5m towards AD.   Speed = dis tance time R  32  4 2   Velocity = displacement time   Unit : In M.K.S. system = ms–1 In C.G.S. system = cm/s Instantaneous speed : The speed of a body at a particular instant of time is called its instantaneous speed. s ds = lim  t  0 t dt   If time distance graph is given then speed can be given by the slope of the line, at given time v= The s = slope t (b) Uniform and Non uniform speed Uniform speed : If an object covers equal distance in equal interval of time, then time speed graph of an object is a straight line parallel to time axis then body is moving with a uniform speed. area of velocity time graph gives Non-uniform speed : displacement travelled. If the speed of a body is changing with respect to time it is moving with a nonuniform speed. Types of speed (a) Average and Instantaneous speed Average speed : It is obtained by dividing the total distance travelled by the total time interval. i.e. average speed = total distance total time Averagevelocity The distance between two points A and B is 100 m. A person moves from A to B with a speed of 20 m/s and from B to A with a speed of 25 m/s. Calculate average speed and average velocity. Sol. (i) Distance from A to B = 100 m Distance from B to A = 100 m Thus, total distance = 200 m Time taken to move from A to B, is given by Ex.3 displaceme nt total time  Average speed is a scalar, while average velocity is a vector.   For a moving body average speed can never be –ve or zero (unless t ), while average t1 = Time taken from B to A, is given by distance 100 t2 = = = 4 seconds velocity 25  velocity can be i.e. vav > 0 while v av > = or < 0  In general average speed is not equal to magnitude of average velocity. However it can be so if the motion is along a straight line without change in direction  If a particle travels distances L1 , L2, L3 at speeds v1, v2, v3 etc respectively, then vav = L  L 2  .....  L n s = 1 = L L L t 1  2  ....  n v1 v 2 vn  Li L v i i  If a particle travels at speeds v1, v2 etc for intervals t1, t2 etc respectively, then vav = v1 t 1  v 2 t 2  .... = t1  t 2  .... v t t 1 1 1 distance 100 = = 5 seconds velocity 20 Total time taken = t1 + t2 = 5 + 4 = 9 sec.  Average speed of the person Total distance covered 200 = = m/s = 22.2 m/s Total time taken 9 (ii) Since person comes back to initial position A, displacement will be zero, resulting zero average velocity. Ex.4 A car moves with a speed of 40 km/hr for first hour, then with a speed of 60 km/hr for next half hour and finally with a speed of 30 1 km/hr for next 1 hours. Calculate the 2 average speed of the car. Sol.  Distance travelled in first hour, is given by s1 = speed × time = 40 km/hr × 1 hr = 40 km Distance travelled in next half an hour, is given by 1 s2 = speed × time = 60 km/hr × hr = 30 km 2 1 Distance travelled in last 1 hours, is given by 2 3 s3 = speed × time = 30 km/hr × hr = 45 km 2 Thus, total distance travelled = s1 + s2 + s3 = 40 + 30 + 45 = 115 km 1 1 Total time taken = 1 + + 1 = 3 hours 2 2  Average speed = Total distance covered 115km = =38.33 km/hr 3hrs Total time taken Figure shows time distance graph of an object. Calculate the following : (i) Which part of the graph shows that the body is at rest ? (ii) Average speed in first 10 s. (iii) Speeds in different parts of motion. Ex.6 Time-velocity graph of a particle is shown in Figure. Calculate the distance travelled in first seconds. C 100 80 A B M 60 40 20 E F G D 0 1 2 3 4 5 6 7 8 9 10 Time(s) Sol. Distance travelled in first 8s is given by area OABCG = area of rectangle OAMG + area of triangle BMC = 8 × 60 + Ex.5 1 × 4 × 40 = 480 + 80 = 560 m. 2 (iii) Acceleration  Rate of change of velocity is called acceleration. It is a vector quantity i.e. a = vu t  Unit of acceleration = m/s2 or ms–2 Distance-time graph Sol. (i) The part BC shows that the body is at rest. (ii) In first 10 seconds, distance travelled = 100m  Thus, average speed = Distance covered Time taken 100 = 10 m/s 10 (iii) Speed of the object in part AB is given by 100 slope = = 50/3 m/s 6 Speed of object in part BC = 0 m/s Speed of the object in part CD 100  40 60 = = = 30 m/s 2 12  10 Speed of object in part DE 40  0 40 = = = 20m/s 14  12 2 = Types of acceleration  Uniform & Non uniform acceleration Uniform acceleration If a body travels in a straight line and its velocity increases by equal amounts in equal intervals of time then it is said to be in state of uniform acceleration. e.g. motion of a freely falling body. Non uniform acceleration A body has a non-uniform acceleration if its velocity increases by unequal amounts in equal intervals of time. Instantaneous acceleration  The acceleration of a body at any instant is called its instantaneous acceleration.  If the velocity of a body decreases, then it will experience a negative acceleration which is called deceleration or retardation.  Acceleration is determined by the slope of time-velocity graph. tan  = Sol. (i) Instantaneous acceleration at t = 3s, is given by a = slope of line AB = zero dv dt (i) If the time velocity graph is a straight line, (ii) Instantaneous acceleration at t = 6 s, is given by a = slope of line acceleration remains constants. (ii) If the slope of the straight line is positive, BC = positive acceleration occurs. (iii) If the slope of the straight line is negative, negative acceleration or retardation occurs. Ex.7 Time-velocity graph of a body is shown in the figure. Find its acceleration in m/s2. As it is clear from the figure, Sol. At t = 0 s, v = 20 m/s At t = 4 s, v = 80 m/s  Acceleration, a = Change in velocity Time int erval = (iii) Instantaneous acceleration at t = 9 s, is given 0  100 = –50 m/s2 by a = slope of line CD = 10  8 Ex.9 Starting from rest, Deepak paddles his bicycle to attain a velocity of 6 m/s in 30 seconds then he applies brakes so that the velocity of the bicycle comes down to 4 m/s in the next 5 seconds. Calculate the acceleration of the bicycle in both the cases. Sol. (i) Initial velocity, u = 0, final velocity, 80 60 40 20 0 CM 100  60 = = – 10 m/s2 BM 84 v = 6 m/s, time, t = 30 s Using the equation v = u + at, we have 1 2 3 4 Time(hrs) v v (80  20) m / s v 15m/s2 = 2 1 = t 2  t1 (4  0) t a= vu t substituting the given values of u, v and t in the above equation, we get Ex.8 Time-velocity graph of a particle is shown in figure. Find its instantaneous acceleration at following intervals : a= 60 = 0.2 m/s2 ; 30 which is positive acceleration. (ii) Initial velocity, u = 6 m/s, final velocity, v = 4 m/s, time, t = 5 s, then a= vu 46 = = –0.4 m/s2 ; t 5 which is retardation. (i) at t = 3s (ii) at t = 6s (iii) at t = 9s Note : The acceleration of the case (i) is positive and is negative in the case (ii).  Motion under uniform acceleration (a) 1st Equation of motion Consider a body having initial velocity 'u'. Suppose it is subjected to a uniform acceleration 'a' so that after time 't' its final velocity becomes 'v'. Now we know, change in velocity Acceleration = Time vu a= t or v = u + at …..(i) (b) 2nd Equation of motion Suppose a body has an initial velocity 'u' and uniform acceleration 'a' for time 't' so that its final velocity becomes 'v'. The distance travelled by moving body in time 't' is 's' then the average velocity = (v + u)/2. Distance travelled = Average velocity × time  u v  u  u  at  t  s =  t (as v = u + at) 2 2     s=  2ut  at 2  2u  at  s=   t s = 2  2  1 s = ut + at 2 ……(ii) 2  The equations of motion under gravity can be obtained by replacing acceleration by acceleration due to gravity (g) and can be written as follows :  When the body is coming towards the centre of earth (a) v = u + gt 1 2 gt 2 (b) h = ut + (c) v2 = u2 + 2gh  When a body is thrown upwards with some initial velocity, then a retardation produced due to attraction of the earth. In equations of motion, a is replaced by (–g) and thus equations become. (a) v = u – gt (b) h = ut – 1 2 gt 2 (c) v2 = u2 – 2gh Assuming u = 0 for a freely falling body : t is given h is given v is given v  gt 2h t g t v g h v2 2g 1 h  gt 2 2 v  2gh (c) 3rd Equation of motion Distance travelled = Average velocity x time uv s=  t  2  ……(iii) from equation (i) t= vu a Substituting the value of t in equation (iii),  Body is projected vertically up : Taking initial position as origin and direction of motion (i.e. vertically up) as positive. (a) At the highest point v = 0 (b) a = – g vu  vu  we get s =     a  2   v2  u 2 s =   2a 2 2 2 1 2      2as = v – u 2h / g 2 2 or v = u + 2as….(iv) 2 2gh u g u2 2g It is clear that in case of motion under gravity (a) Time taken to go up is equal to the time taken to fall down through the same distance. (b) The speed with which a body is projected up is equal to the speed with which it comes back to the point of projection. (c) The body returns to the starting point with the same speed with which it was thrown. Ex.12 A body starts from rest and moves with a constant acceleration. It travels a distance s1in first 10 s, and a distance s2 in next 10 s. Find the relation between s2 and s1. Given : u = 0, t1 = 10 s Sol.  Distance travelled in first 10 seconds, is given by 1 2 1 at = 0 + × a × (10)2 2 2 = 50a ...(1) s1 = ut + Ex.10 A body starts moving with an initial velocity 50 m/s and acceleration 20 m/s2. How much distance it will cover in 4s ? Also, calculate its average speed during this time interval. Sol. To calculate the distance travelled in next 10s, we first calculate distance travelled in 20 s and then subtract distance travelled in first 10 s. s = ut + Given : u = 50 m/s, a = 20 m/s2, = 200a t = 4s, s = ? s = ut + 1 2 1 at = 50 × 4 + × 20 × (4)2 2 2 1 2 1 at = 0 + × a × (20)2 2 2 ...(2)  Distance travelled in 10th second interval, s2 = s – s1 = 200a – 50a ...(3) or s2 = 150a = 200 + 160 = 360 m Now, s2 150a 3 = = s1 1 50a or s2 = 3s1 Average speed during this interval, distance travelled 360 V = = = 90 m/s 4 time interval Ex.11 A body is moving with a speed of 20 m/s. When certain force is applied, an acceleration of 4 m/s2 is produced. After how much time its velocity will be 80 m/s ? Sol. Given : u = 20 m/s, a = 4 m/s2, Ex.13 A train is moving with a velocity 400 m/s. With the application of brakes a retardation of 10 m/s2 is produced. Calculate the following : (i) After how much time it will stop ? (ii) How much distance will it travel before it stops? Sol. (i) Given: u = 400 m/s, a = –10 m/s2, v = 0, t = ? Using equation, v = u + at, we get v = 80 m/s, t = ? 0 = 400 + (–10) × t Using equation, v = u + at, we get 80 = 20 + 4 × t or t = 40 s (ii) For calculating the distance travelled, we use equation, or 4t = 80 – 20 = 60 v2 = u2 + 2as, we get or t = 15 s (0)2 = (400)2 + 2 × (–10) × s Therefore, after 15 seconds, the velocity of the body will be 80 m/s. or or 20s = 400 × 400 s = 8000 m = 8 km Ex.14 A body is thrown vertically upwards with an initial velocity of 19.6 m/s. If g = –9.8 m/s2. Calculate the following : (i) The maximum height attained by the body. (ii) After how much time will it come back to the ground ? Sol.(i) Given: u = 19.6 m/s, g = –9.8 m/s2, v = 0, h = ? Using equation v2 = u2 + 2gh, we get (0)2 = (19.6)2 + 2(–9.8) × h 19.6 19.6 or h= = 19.6 m 2  9.8 (ii) Time taken to reach the maximum height can be calculated by the equation, v = u + gt or 0 = 19.6 + (–9.8) × t or t = 2s In the same time, it will come back to its original position.   Total time = 2 × 2 = 4s Ex.15 From the top of a tower of height 490 m, a shell is fired horizontally with a velocity 100 m/s. At what distance from the bottom of the tower, the shell will hit the ground ? Sol. We know that the horizontal motion and the vertical motion are independent of each other. Now for vertical motion, we have u = 0, h = 490 m, g = 9.8 m/s2, t = ? 1 Using equation,h = ut + gt2, we get 2 1 490 = 0 + × 9.8 × t2 2 490 or t2 = = 100 4.9 or t = 10 s  It takes 10 seconds to reach the ground. Now, horizontal distance = horizontal velocity × time = 100 m/s × 10 s = 1000 m  The shell will strike the ground at a distance of 100 m from the bottom of the tower. Displacement- time graph :  The straight line inclined to time axis in s-t graph represents constant velocity.  In s-t graph the straight line inclined to time axis at angle greater than 90º shows negative velocity  Body with accelerated motion  Body with decelerated motion Velocity -time graph :  For the body having constant velocity or zero acceleration. v t  The body is moving with constant retardation and its initial velocity is not zero.  The body is accelerated and the initial velocity is zero.  The body is decelerated Acceleration-time graph :  Acceleration is constant  Acceleration is increasing and is +ve Acceleration is decreasing and is –ve Centripetal acceleration When a body moves in such a way that its distance from a fixed point always remains constant, then its motion is said to be the circular motion.  In uniform circular motion the particle experiences an acceleration called the centripetal acceleration.  ac = Uniform circular motion :  If the radius vector sweeps out equal angles in equal times, then its motion is said to be uniform circular motion. v2 r  The direction of centripetal acceleration is along the radius towards the centre. Centripetal force :  Always acts towards centre.  In uniform circular motion speed remains const.  Linear velocity, being a vector quantity, its direction changes continuously.  The direction of velocity is along the tangent at every point.  Because Fc is always perpendicular to velocity or displacement, hence the work done by this force will always be zero. Note :  Circular motion in horizontal plane is usually uniform circular motion. Angular velocity : =  Centripetal force is required to move a particle in a circle.  t  Remember that equations of motion are not applicable for circular motion.  A vector quantity  Direction is perpendicular to plane of rotation Note : If the particle is revolving in the clockwise direction then the direction of angular velocity is perpendicular to the plane downwards. Whereas in case of anticlockwise direction, the direction will be upwards.  Unit is Radian/sec.  v = rwhere r = radius of the circle.  It is the time taken to complete one complete revolution.  In one revolution, angle subtended is 2 and if T is time period, then the angular velocity is given by =  In uniform circular motion the direction of angular velocity is along the axis of rotation which is constant throughout.  Angular velocity remains constant magnitude as well as in direction. Time period : in 2 T or T= 2  Frequency :  Frequency is defined as the number of revolutions per second. i.e. n = 1  = T 2 Ex.16 A particle moves in a circle of radius 2 m and completes 5 revolutions in 10 seconds. Calculate the following : (i) Angular velocity and (ii) Linear velocity. Since, it completes 5 revolutions in 10 seconds. Sol.   Time period = 10 = 2s 5 (i) Now angular velocity, = 2 T = 2 = 2 rad/s (ii) Linear velocity is given by v = r= 2   v = 2m/s Ex.17 The length of second's needle in a watch is 1.2 cm. Calculate the following : (i) Angular velocity and (ii) Linear velocity of the tip of the needle. Sol. (i) We know that the second's needle in a watch completes one revolution in 60 seconds.   Time period, T = 60 s Angular velocity, 2 2     = = = rad/s T 60 30 (ii) Length of the needle = 1.2 cm = Radius of the circle Linear velocity of the tip of the needle is given by   v = r= 1.2 × = 30 25 or v =  = 1.266 × 10–1 cm/sec. 2s Ex.18 Earth revolves around the sun in 365 days. Calculate its angular velocity. Sol. Time period, T = 365 days = 365 × 24 × 60 × 60 seconds  Angular velocity, = = 2 T 2 rad/s = 1.99 × 10–7 rad/s. 365  24  60  60