MOTION IN A STRAIGHT LINE Distance It is equal to the actual path traversed by an object between the initial and final positions. It is a scalar quantity. It is always positive. In the given figure it is equal to ACB if a particle moves from A to B via C. Displacement The shortest distance between the initial and final positions of an object is the magnitude of displacement and the vector drawn from initial position to final position is the displacement vector. In the given figure it is equal to AB if a particle moves from A to B via C. Displacement may be positive, negative or zero. Exercise 1 An athlete runs on a semicircular track of radius 20 m. If he starts at one end of the track and reaches the other end, find the distance covered and the displacement of the athlete. Equations of Motion In One Dimension The first and the second equations of motion are vector equation while the third one is scalar equation as given below: ®v = ®u + ®a t _ _ _ (1) ®s = ®u t + ½ ®a t 2 _ _ _ (2) and ® ®v . v = ® ®u .u + 2 ® ®a . s _ _ _ (3) Equations (1) and (2) give the values of velocity (v) and displacement (s) after time t. There is no need to use vector notation for these two equations if ®u and ®a are in the same direction. If the initial velocity ®u and acceleration ®a of an object make an angle q with each other, then the magnitude and direction of the final velocity can be calculated from the law of vector addition. NOTE 1. The distance S n during the th n second is given by S n = u + a2 (2n – 1) where u is initial velocity and a is acceleration. 2. The ratio of the distance covered in the n th second to the distance covered in n seconds by a body starting from rest and moving with constant acceleration is given by SSnnth = 2n – 1 n2 Exercise 2 An object starting from rest and moving with constant acceleration covers 10 m in the first second. Find the distance covered by it in the third second. NOTE The numerical ratio of displacement to distance is £ 1. The value of this ratio is equal to 1 in a uniform motion along a straight line in a given direction. Average Speed If Ds be the distance travelled in the time interval Dt, then the average speed in this time interval is v av = DDst The instantaneous speed at a time t is defined as v = lim Dt ®0 DDst = ds dt NOTE 1. If an object travels along a straight line with speed v1 for the first half distance and then with speed v 2 for the remaining half distance , then its average speed during the whole journey is given by V av = 2v v v v 1 2 1 2 + (using V av = Total time Total distance ) Here, V av is equal to the harmonic mean of v1 and v 2 . 2. If an object travels along a straight line with speed v1 for time t 1 and with speed v 2 for time t 2 , the average speed of the object during the whole journey is given by V av = 1 2 1 1 2 2 t t v t v t ++ (using V av = Total time Total distance ) If t 1 = t 2 , V av = v v 1 2 2+ Here, V av is equal to the arithmetic mean of v1 and v 2 . Exercise 3 A car has to cover a distance of 60 km. If half of the total time it travels with a speed of 80 km/h and in rest half time, its speed becomes 40 km/h. What is the average speed of the car? 3. If an object covers first one-third distance with speed v1 , next one-third with v 2 and last one-third with 3 v , then average speed is given V av = 1 2 2 3 3 1 1 2 3 3 v v v v v v v v v+ + (using V av = Total time Total distance ) Exercise 4 A bus traveled the first one-third distance at the speed of 10 km/h, the next one-third at 20 km/h and the last one-third at 60 km/h. What is the average speed of the bus? 4. If an object covers first one-third of total distance with speed v1 and the remaining part with a speed v 2 for half the time and with 3 v for the other half of the time, then the average speed of the object during the whole journey is given by V av = 1 2 3 1 2 3 4 3 ( ) v v v v v v + + + (using V av = Total time Total distance ) 5. Initial velocity is taken equal to zero, i.e., u = 0 if an object starts from rest or falls freely. 6. If the stopping distance for an object of mass m moving with speed v is s, then for a speed nv, the stopping distance becomes n 2 s (for the same retardation). Exercise 5 A car moving at a speed of 10 m/s is stopped over a distance of 30 m after brakes are applied. What will be the distance if it is moving at 30 m/s? Exercise 6 A rifle bullet loses 1/20th of its velocity in passing through a plank. What is the least number of such planks required just to stop the bullet? Velocity If D s® be the displacement travelled in the time interval Dt, then the average velocity in this time interval is given by vav ® = DDst® The instantaneous velocity at a time t is defined as v® = lim Dt ®0 DDst® = d s dt® Exercise 7 A car moves along a straight line whose motion is given by s = 12 2 t − 4t + 6, where s is in metres and t is in seconds. What is its initial velocity? NOTE 1. In uniform motion, the average and instantaneous velocities are always same the velocity is same at each point of its path or at each instant. 2. In uniform motion, the velocity is independent of the choice of time interval and origin. 3. Graphically both the displacement-time and velocity-time graphs can represent a uniform motion. The lines OC and DE in the displacement-time graph and line AB in the velocity-time graph (next articles) represent uniform motion. 4. Two ends of a train moving with constant acceleration pass a certain point with velocities v1 and v 2 , then its middle point passes that point with the velocity V = v v 12 22 2+ Exercise 8 An object moving with a uniform acceleration has velocities of 30 m/s and 40 m/s when passing points A and B in its path. What is the velocity midway between A and B? 5. Two trains are moving on the same track in the same direction with velocities v1 (running ahead of the two) and v 2 such that v 2 > v1 . If a retardation a is produced in the second train, then minimum time to avoid collision is t = v v a 2 1 − 6. Two trains are moving on the same track in the same direction with velocities v1 (running ahead of the two by a distance d) and v 2 such that v 2 > v1 . If a retardation a is produced in the second train, then to avoid collision d > (v v ) a 2 1 2 2− NOTE 1. Average speed of a particle in a given time is never less than the magnitude of the average velocity. 2. It is possible to have a situation in which d v dt® ¹ 0 but ddt v® = 0. 3. If the average velocity of a particle is zero in a time interval, then it is possible that the instantaneous velocity is never zero in that interval. If a particle comes back to initial position, the average velocity is zero but the instantaneous velocity not zero. 4. A particle can have acceleration without having any velocity. For example, a particle thrown vertically up has zero velocity at the highest point but a downward acceleration. A simple pendulum at the extreme position also has this property. 5. An object can have constant speed but variable velocity (as in uniform circular motion) but the opposite is not true, i.e., constant velocity with variable speed is not possible. Displacement-Time Graph 1. The graph is parallel to t-axis. There is no change in displacement with time (s µ 0 t ). It shows that v = 0 (line AB). 2. The graph is an oblique straight line. This line has constant slope, i.e., velocity is constant (line OC and DE). There is no acceleration in the motion (as s µ t, and, hence, v µ 0 t ). So, no force is required to make the object move with uniform velocity. In this case, instantaneous velocity is always equal to the average velocity. 3. DE cannot represent distance-time graph as it means decreasing distance with increasing time (i.e., negative slope). It is not possible as distance always increases with time. OC can represent both displacement-time and distance-time graphs. 4. The displacement-time graph is a curve. Curve OB -velocity increases with time as the slope of the curve increases. Motion is accelerated. To have constant acceleration, s µ 2 t (a parabola) and hence, v µ t (a straight line). This graph is a plot of the equation s = ut + ½ g 2 t . Curve OA-velocity decreases with time as the slope decreases. Motion is decelerated. This graph is a plot of equation s = ut − ½ g 2 t . 5. If s µ 3 t and hence, v µ 2 t , then a µ t. In this case, the acceleration is not constant but varies with time. NOTE 1. Displacement-time curve cannot be parallel to displacement axis as it means a finite displacement in no time. In other words, it denotes an infinite velocity. 2. Displacement-time and distance-time graph cannot be a closed curve in one dimensional motion as it means that the object will be at two positions at the same instant of time. Also, the reversal of time is not possible. 3. The displacement-time graph cannot take sharp turns as it gives two different velocities at that point. 4. The displacement-time graph cannot be symmetric about the time-axis as a particle cannot have two displacements at an instant. But this graph may be symmetric about the displacement-axis. 5. The displacement-time graph does not show the trajectory of the object. 6. The distance-time graph is always an increasing curve for a moving object as distance never decreases with time. 7. The distances covered by an object, starting from rest, in successive seconds are in the ratio 1 : 3 : 5 ---etc. But the distances covered in the first second, first two seconds, first three seconds, ---etc. are in the ratio 1 : 4 : 9 : 16 : ---etc. Exercise 9 Figure shows the displacement-time graph of a particle moving along a straight line. Find (a) the average velocity during 0 to 10 s, (b) instantaneous velocity at 2, 5, 8 and 12 s. Velocity-Time Graph 1. The v-t graph is parallel to time axis. The velocity constant and hence, acceleration = 0 (AB). This line on v-t graph is equivalent to lines OC and DE on s–t graph. 2. The graph is an oblique straight line. The motion in this case has constant acceleration (OC) or deceleration (DE) as velocity changes at constant rate. 3. The graph is a curve of increasing slope. It shows that the object has variable velocity which is increasing with time. The motion is accelerated (OB) and the acceleration goes on increasing. 4. The graph is a curve of decreasing slope. It shows that the object has variable velocity which is decreasing with time. The motion has decreasing acceleration (OA). 5. Area between v-t graph and time-axis gives the displacement in that time interval along with appropriate sign. 6. The distance covered by a body in a given time interval is equal to total area between v-t graph and time-axis without considering sign. 7. With help of a given v-t graph, both s-t graph and a-t graph can be plotted. 8. A v-t graph cannot be a closed curve in one-dimensional motion. The following v-t graph is not possible in one dimension as the particle will have velocities in positive as well as negative directions. 9. The velocity-time graph can take sharp turns. 10. A speed-time graph as shown below is not possible as the speed of an object can never be negative. Exercise 10 Figure shows the velocity-time graph of a particle moving along a straight line. Find (a) the acceleration, (b) the distance travelled during 0 to 10 s, (c) the displacement in 0 to 10 s (d). Exercise 11 Figure shows the time-displacement graph for a particle in one dimension. Draw the corresponding time-velocity graph. Acceleration-Time Graph 1. The graph is a line parallel to time axis. The acceleration is constant (AB) 2. The graph is an oblique straight line. The acceleration is uniformly increasing (OC) or decreasing (DE). 3. The area under a-t graph and time axis gives the change in velocity of the object in that time interval. 4. With help of a given a-t graph, v-t graph can be plotted. 5. Acceleration-time graph cannot be a closed curve. 6. If an object travels along a straight line with uniform acceleration a1 for time t 1 and with uniform acceleration a 2 for time t 2 , the average acceleration of the object during the whole journey is given by a av = 1 2 1 1 2 2 t t a t a t ++ If t 1 = t 2 , a av = 2 1 2 a + a Vertical Motion Under Gravity 1. If a body is thrown upward with velocity u, then the maximum height H is H = ug2 2 [Putting v = 0, s = H and a = − g in v 2 = u 2 + 2as] 2. If a body is thrown upward with velocity u, then the time in which it reaches the maximum height is t = ug [Putting v = 0, and a = − g in v = u + at] 3. If a body is dropped from a height h, then the velocity when it reaches the ground is v = 2gh [Putting u = 0, s = h and a = g in v 2 = u 2 + 2as] 4. If a body is dropped from a height h, then the time to reach the ground is t = 2h g [Putting u = 0, s = h and a = g in s = ut + ½ at 2 ] 5. A body thrown upward with velocity u is at the same height h at two instants of time (let t 1 and t 2 ). Then, h = 12 1 2 gt t and u = 12 1 2 g(t + t ) The time taken to reach the highest point is t = 12 1 2 (t + t ) 6. If an object is dropped at a height h from a balloon moving upward with velocity u, then the time taken by the object to reach the ground can be calculated from the relation h = − ut + ½ gt 2 It must be noted that at the instant of releasing the object, its velocity is equal to that of the balloon at that instant and is upward (taken negative) whereas its displacement and acceleration are downward (taken positive). 7. If three balls are thrown with same speed such that one is thrown vertically downward, the second vertically upward and the third horizontally, then all the three balls reach the ground with same speed. If t 1 , t 2 and t 3 are the respective time taken by the balls to the reach the ground, then t 3 = 1 2 t t 8. If retarding force due to air resistance is considered, then an object takes lesser time to reach the highest point and larger time to reach the ground as compared to that taken in the absence of air resistance. In the first case, effective value of g increases as air resistance acts downward. In the second case, effective value of g decreases as air resistance acts upward. Exercise 12 A balloon is rising vertically upwards with uniform acceleration 15.7 m/s 2 . A stone is dropped from it. After 4 sec another stone is dropped from it. Find the distance between the two stones 6 sec after the second stone is dropped. Exercise 13 A balloon is rising vertically upwards with uniform speed of 10 m/s. When it is 400 m above the ground, a stone is dropped from it. After how much time and with what velocity will the stone hit the ground? (g = 10 m/s 2 ) Answers: 1. 20p m, 40 m 2. 50 m 3. 60 km/h 4. 18 km/h 5. 270 m 6. 11 7. 24 m/s 8. 25 2 m 9. (a) 10 m/s (b) 20 m/s, 0 m/s, 20 m/s, − 20 m/s 10. (a) 0.6 m/s 2 (b) 50 m (c) 50 m 12. 816 m 13. 10 sec, 90 m/s