Chapter – 2
Motion in a straight line
Q 1. In which of the following examples of motion can the body be consider dapp rox ima tely a point object?
(a) A railway carriage moving without jerks between two stations.
(b) A cap on top of a man cycling smoothly on a circular track.
(c) A spinning cricket ball that turns sharply on hitting the ground.
(d) A tumbling beaker that has slipped off the edge of a table.
Ans.
(a), (b) The size of the railway carriage and the cap is very small as compared to the distance they’ve travelled, i.e. the distance between the two stations and the length of the race track, respectively. Therefore, the cap and the carriage can be considered as point objects.The size of the cricket ball is comparable to the distance through which it bounces off after hitting the floor. Thus, the cricket ball cannot be treated as a point object. Likewise, the size of the beaker is comparable to the height of the table from which it drops. Thus, the beaker cannot be treated as a point object.
Q2. The position-time (x-t) graphs for two children, A and B, returning from their school O to their homes, P and Q, respectively, are shown in Fig. Choose the correct entries in the brackets below:
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time(e) (A/B) overtakes (B/A) on the road (once/twice).Motion in a straight line image 1
Ans.(a) A lives closer to the school than B, because A has to cover shorter distances [OP < OQ].
(b) A starts from school earlier than B, because for x= 0, t = 0 for A but for B, t has some finite time.
(c) The slope of B is greater than that of A; therefore, B walks faster than A.
(d) Both A and B will reach their home at the same time.
(e) At the point of intersection, B overtakes A on the roads once.
Q3. A woman starts from her home at 9.00 am, walks at a speed of 5 km/h on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by auto at a speed of 25 km/h. Choose suitable scales and plot the x-t graph of her motion.Motion in a straight line Image 2
Ans.
Distance to her office = 2.5 km.Walking speed the woman = 5 km/hTime taken to reach office while walking = (2.5/5 ) h=(1/2) h = 30 minutesSpeed of auto = 25 km/hTime taken to reach home in auto = 2.5/25 = (1/10) h = 0.1 h = 6 minutesIn the graph, O is taken as the origin of the distance and the time, then at t = 9.00 am, x = 0and at t = 9.30 am, x = 2.5 kmOA is the portion on the x-t graph that represents her walk from home to the office. AB represents her time of stay in the office from 9.30 to 5. Her return journey is represented by BC.
Q4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backwards, followed again by 5 steps forward and 3 steps backwards, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.Motion in a straight line Image 3
Ans.
The time taken to go one step is 1 second. In 5 s, he moves forward through a distance of 5 m, and then in the next 3 s, he comes back by 3 m. Therefore, in 8 s, he covers 2 m. So, to cover a distance of 8 m, he takes 32 s. He must take another 5 steps forward to fall into the pit. So, the total time taken is 32 s + 5 s = 37 s to fall into a pit 13 m away
Q5. A jet aeroplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Ans.
Speed of the jet aeroplane, VA= 500 km/hThe speed at which the combustion products are ejected relative to the jet plane, VB – VA= – 1500 km/h(The negative sign indicates that the combustion products move in a direction opposite to that of the jet)Speed of combustion products w.r.t. observer on the ground, VB – 500 = – 1500VB = – 1500 + 500 = – 1000 km/h
Q6. A car moving along a straight highway with a speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Ans.
The initial velocity of the car = uThe final velocity of the car = vDistance covered by the car before coming to rest = 200 mUsing the equation,v = u + att = (v – u)/a = 11.44 sec.Therefore, it takes 11.44 sec for the car to stop.
Q7. Two trains, A and B, of length 400 m each, are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver ofB decides to overtake A and accelerates by 1 m s–2. If, after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Ans.
Length of trains A and B = 400 m Speed of both the trains = 72 km/h = 72 x (5/18) = 20 m/s Using the relation, s = ut + (1/2)at 2 Distance covered by the train BSB = uBt + (1/2)at 2 Acceleration, a = 1 m/sTime = 50 sSB = (20 x 50) + (1/2) x 1 x (50)2= 2250 m Distance covered by the train ASA = uAt + (1/2)at 2 Acceleration, a = 0SA = uAt = 20 x 50 = 1000 m Therefore, the original distance between the two trains = SB – SA = 2250 – 1000 = 1250 m
Q8. On a two-lane road, car A is travelling at a speed of 36 km/h. Two cars, B andC, approach car A in opposite directions with a speed of 54 km/h each. At acert a in instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What is the minimum acceleration of car B required to avoid an accident?
Ans:
The speed of car A = 36 km/h = 36 x (5/8) = 10 m/s Speed of car B = 54 km/h = 54 x (5/18) = 15 m/s Speed of car C = – 54 km/h = -54 x (5/18) = -15 m/s (negative sign shows B and C are in opposite directions)Relative speed of A w.r.t. C, VAC = VA – VB = 10 – (-15) = 25 m/s Relative speed of B w.r.t. A, VBA = VB – VA = 15 – 10 = 5 m/s Distance between AB = Distance between AC = 1 km = 1000 mTime taken by the car C to cover the distance AC, t = 1000/VAC = 1000/ 25 = 40 sIf a is the acceleration, thens = ut + (1/2) at21000 = (5 x 40) + (1/2) a (40) 2 a = (1000 – 200)/ 800 = 1 m/s 2 Thus, the minimum acceleration of car B required to avoid an accident is 1 m/s 2
Q9. Two towns, A and B, are connected by regular bus service, with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h–1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion and every 6 min in the opposite direction. What is the period T of the bus service, and with what speed (assumed constant) do the buses ply on the road?
Ans.
Speed of each bus = Vb Speed of the cyclist = Vc = 20 km/hThe relative velocity of the buses plying in the direction of motion of the cyclist is Vb – Vc.The buses playing in the direction of motion of the cyclist go past him after every 18 minutes, i.e. (18/60) s.Distance covered = (Vb – Vc ) x 18/60 Since the buses are leaving every T minutes. Therefore, the distance is equal to Vb x (T/60)(Vb – Vc ) x 18/60 = Vb x (T/60) ——(1)The relative velocity of the buses plying in the direction opposite to the motion of the cyclist is Vb + Vc The buses go past the cyclist every 6 minutes, i.e. (6/60) s.Distance covered = (Vb + Vc ) x 6/60Therefore, (Vb +Vc ) x 6/60 = Vb x (T/60)——(2)Dividing (2) by (1)[(Vb – Vc ) x 18/60]/ [(Vb + Vc ) x 6/60 ]= [Vb x (T/60)] /[Vb x (T/60)](Vb – Vc ) 18/(Vb +Vc ) 6 = 1(Vb – Vc )3 = (Vb +Vc )Substituting the value of Vc(Vb – 20 )3= (Vb + 20 )3Vb – 60 = Vb + 202Vb = 80Vb = 80/2 = 40 km/hTo find the value of T, substitute the values of Vb and Vc in equation (1)(Vb – Vc ) x 18/60 = Vb x (T/60)(40 – 20) x (18/60) = 40 x (T/60)T = (20 x 18) /40 = 9 minutes
Q10. A player throws a ball upwards with an initial speed of 29.4 m/s.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of the x-axis, and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise, and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).
Ans.
(a) The acceleration due to gravity always acts downwards towards the centre of the Earth.(b) At the highest point of its motion, the velocity of the ball will be zero, but the acceleration due to gravity will be 9.8 m s–2 acting vertically downward.
(c) If we consider the highest point of ball motion as x = 0, t = 0, and vertically downward direction to be +ve direction of the x-axis, then
(i) During the upward motion of the ball before reaching the highest point position, x = +ve, velocity, v = -ve and acceleration, a = +ve.
(ii) During the downward motion of the ball after reaching the highest point position, velocity and acceleration, all three quantities are positive.(d) Initial speed of the ball, u= -29.4 m/sThe final velocity of the ball, v = 0 Acceleration = 9.8 m/s 2 Applying in the equation v2 – u2 = 2 gs 0 – (-29.4)2 = 2 (9.8) ss = – 864.36/19.6 = – 44.1Height to which the ball rises = – 44.1 m (negative sign represents upward direction)Considering the equation of motion v = u + at 0 = (-29.4) + 9.8tt = 29.4/9.8 = 3 seconds Therefore, the total time taken for the ball to return to the player’s hands is 3 +3 = 6 s.
Q11. Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity
(c) with constant speed must have zero acceleration
(d) with positive value of acceleration must be speeding up
Ans.
(a) True
(b) False
(c) True (if the particle rebounds instantly with the same speed, it implies infinite acceleration, which is unphysical)
(d) False (true only when the chosen positive direction is along the direction of motion)
Q 12. A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Ans.
Height from which the ball is dropped = 90 mThe initial velocity of the ball, u = 0 Let v be the final velocity of the ball Using the equation v 2 – u2 = 2 as ——–(1)v 12 – 0 = 2 x 10 x 90 v 1= 42.43 m/sTime taken for the first collision can be given by the equation v = u + at42.43 = 0 + (10) tt 1 = 4.24 sThe ball losses one-tenth of the velocity at collision. So, the rebound velocity of the ball is v 2= v – (1/10)vv 2 = (9/10) vv 2= (9/10) (42.43)= 38.19 m/sTime taken to reach maximum height after the first collision isv = u + at38.19 = 0 + (10)t 2 t 2 = 3.819 sThe total time taken by the ball to reach the maximum height isT = t 1 + t 2 T = 4.24 + 3.819 = 8.05 sNow the ball will travel back to the ground in the same time as it took to reach the maximum height = 3.819 s Total time taken will be, T = 4.24 + 3.819 + 3.819 = 11.86Velocity after the second collision v 3 = (9/10) (38.19)v3 = 34.37 m/s Using the above information, the speed time graph can be plotted Motion in a straight line Image 4