|Motion in straight line!|
NCERT Chapter Solutions
Q3.1 In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting 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.
Answer: The objects in motion can be considered as point objects, so far as the size of the object is much smaller than the distance it moves in a reasonable duration of time.
(a) The size of railway carriage is very small when compared to the distance between stations. It can be considered as point object.
(b) Size of monkey in comparison to circular track is small. Therefore we can consider monkey as point object.
(c) The size of the cricket ball as compared to the turning path while hitting the ground is not negligible. In this case, we cannot consider ball as point object.
(d) Size of tumbling beaker as compared to length of the edge is not small. Here, the beaker cannot be treated as point object.
Q 3.2. 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. 3.19. 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).
Answer: Let us extend A and B line segments. Draw P & Q lines parallel to time-axis, as shown in fig-A).
(a) A lives closer to school than B. (since OP is shorter than OQ).
(b) A starts from the school earlier than B. (A starts when t = 0).
(c) B walks faster than A. (Slope of B is higher than the slope A).
(d) As per the Fig-A, it is obvious from the figure, A and B reach home at different time.
(Note: NCERT book says both children reach at the same time. In that case, the diagram should be like Figure-B. To support your answer, you must draw a rough graph.)
(e) B overtakes A on the road once.
Q3.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km/hr 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 an auto with a speed of 25 km/hr. Choose suitable scales and plot the x-t graph of her motion.
Given, Distance between home and office = 2.5 km
Speed of the woman = 5 km/hr
∴ Time taken to reach office = Distance / Speed = 2.5/5 = 0.5 Hrs = 30 minutes.
⇒ The woman starts from home at 9.00 am and reaches office at9.30 am
Speed of auto = 25 km/hr
∴ Time taken to reach home by auto = Distance / Speed = 2.5 / 25 = 0.1 Hrs = 6 minutes
⇒ The woman will reach home by 5.06 pm
Her motion will be represented on x-t (displacement-time) graph as:
Q 3.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, 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.
Answer: Distance covered in 1s = 1m
Distance covered 5 steps = 5m (takes 5 seconds)
Distance covered in 3 steps (backwards) = 3m (time taken = 3s)
Total Displacement in 8s = 5 - 3 = 2m
⇒ Displacement in 16s = 4m
Displacement in 32s = 8m
In next 5s, he will cover 5m distance and total displacement = 8 + 5 = 13mand the drunkard will fall into pit.
Total time taken = 32s + 5s = 37s.
The displacement-time (x-t) graph will be:
Q3.5 A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?
Answer: Velocity of jet (Vj) = 500 km/hr
Let Velocity of observer on ground (Vo)= 0 km/hr
Let Vc is velocity of combustion.. (Note relative velocity of combustion wrt jet is given, not its actual speed).
Relative velocity of jet w.r.t. to ground = (Vj - Vo) = 500 km/hr
Relative velocity of combustion wrt jet = (Vc - Vj) = -1500 km/hr
Relative speed of combustion w.r.t ground = Vc - Vo = (Vc - Vj) + (Vj - Vo)
= -1500 + 500 = -1000 km/hr ... (answer)
The -ve sign indicates the relative speed of combustion is opposite to the direction of jet.
Q3.6: A car moving along a straight highway with speed of 126 km/h 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 ?
Given, initial velocity (u) of car = 126 km/hr = 126 x 1000/3600 = 35 m/s
Final velocity (v) = 0 m/s
Distance covered (S) = 200m
Using equation v2 - u2 = 2aS
a = (v2 - u2)/2S = (0 - 352)/(2 x 200) = -1225/400 = -3.0625 m/s2 ...(answer)
-ve sign indicates that it is retardation.
Using equation, v = u + at
t = (v-u)/a = -35/-3.06 = 11.44s ... (answer)
Q3.7: Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km/h in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms-2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them ?
For Train A:
Speed = 72 km/h = 72 x 1000/3600 = 20 m/s
acceleration (a) = 0
Time (t) = 50s
Distance covered by A = SA = ut + ½(at2) =20x50 + 0 = 1000m
For Train B:
Initial Speed = 72 km/h = 20 m/s
acceleration (a) = 1 ms-2
Time (t) = 50s
Distance covered by B = SB = ut + ½(at2) = 20x50 + 1x502/2 = 1000 + 1250 = 2250m
The original distance between A and B = 2250 - 1000 = 1250m
The relative velocity of B w.r.t A = VBA = VB - VA = 72 -72 = 0
acceleration (a) = 1 ms-2
Time (t) = 50s
Distance covered by B wrt A= SBA = VBAt + ½(at2) = 0 + 1x502/2 = 1250m
Q3.8: On a two-lane road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?
Velocity of Car A (VA) = 36 km/h = 36x1000/3600 = 10m/s
Velocity of Car B (VB) = 54 km/h = 15m/s
Velocity of Car C (VC) = -15m/s
Relative Speed of B wrt A (VBA) = VB - VA = 15 - 10 = 5m/s
Relative Speed of C wrt A (VCA) = VC - VA = -15 - (10) = -25m/s
∵ AB = AC = 1km = 1000m
Time taken by Car C to cover distance AC,
S = ut + ½(at2)
⇒ -1000 = -VCA.t + 0 = 25t ⇒ t = 40s
⇒ Car B must cover distance within 40s.
∴ Using eqn S = ut + ½(at2) ,
1000 = VBA.t + ½(at2) = 5x40 + ½ax1600
1000 = 200 + 800a ⇒ 800a = 800
⇒ a = 1 ms-2
Q3.9: Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km/h 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?
Explanation: Before solving the problem, let us try to understand the information given.
Let the speed of the bus = Vb km/h
Let the speed of the cyclist = Vc km /h = 20 km/h
The bus leaves at regular intervals of T minutes. Since the buses are in uniform motion. They maintain equal distances of T intervals.
Consider the Ist case when the cyclist moves from Town A to Town B. If the first bus passes the cyclist at point Q, the next bus he will see after 18 minutes i.e. at point R.
⇒ PQ is the distance covered by bus in T minutes = Vb.T
⇒ The distance covered by bus in 18 mins i.e. PR = Vb.18
The cyclist will cover distance QR in 18 mins = Vc.18
Since, PR = PQ + QR
⇒ Vb.18 = Vb.T + Vc.18
⇒ Vb.18 - Vc.18 = Vb.T
or (Vb - Vc).18 = Vb.T
Converting minutes into hours,
or (Vb - Vc).18/60 = Vb.T/60
⇒ (Vb - Vc).18 = Vb.T ...(I)
But (Vb - Vc) is the relative speed of bus wrt to speed of the cycle. It means distance covered by the bus in T minutes is relatively equal to the distance covered by it in 18 mins w.r.t. cycle.
Similarly we can solve the second case, when the cyclist travels from B to A.
Here the relative speed of the bus = (Vb + Vc)
The distance covered by bus in relative speed wrt to cycle in 6 minutes is equal to the distance covered in T minutes.
i.e. (Vb + Vc).6/60 = Vb.T/60 ...(II)
Dividing equation II by I, we get
(Vb + Vc)/ (Vb - Vc) = 3
⇒ (Vb + Vc) = (Vb - Vc). 3
⇒ Vb + 20 = 3Vb - 60
⇒ 2Vb = 80
⇒ Vb = 40 km /h
Put the value of Vb in equation I,
(40 - 20).18 = 40.T
⇒ T = 9 min
Q 3.10: 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 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 ms-2 and neglect air resistance).
(b) Zero. When ball reaches at the maximum height, its velocity will become zero and it starts to fall.
(c) When ball moves upward, the position remains positive. velocity is negative and acceleration acting downwards is positive.
When ball moves downward, the position remains positive, velocity is positive because ball moves downwards. Acceleration is positive (acting downwards).
(Note: When can we take position of ball as -ve? When it crosses its maximum height (taken in figure as x = 0)
(d) Initial velocity u = 29.4 m/s, Final velocity (v) = 0
Using equation v2 - u2 = 2aS
0 - (29.4)2 = -2gH
⇒ H = (29.4 29.4) / (2 x 9.8) = 44.1 m
Using equation v = u + at , where t = time of ascent.
⇒ 0 = 29.4 - 9.8t
⇒ t = 29.4/9.8 = 3s
∵ time of ascent = time of descent
⇒ Total time = 2 × 3 = 6s
Q3.11 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.
(a) with zero speed at an instant may have non-zero acceleration at that instant - ✓ (True)
e.g. A stone is thrown upwards, it reaches maximum height. At that point its speed is zero but the non-zero constant acceleration due to gravity (g) is being applied.
(b) with zero speed may have non-zero velocity, ✗ (False)
The magnitude of velocity is speed. You cannot have non-zero velocity but zero speed. You cannot say a car is not moving but it is going to east.
(c) with constant speed must have zero acceleration ✓ (True)
Consider the equation, v = u + at. In case of constant speed, v = u
∴ u = u + at ⇒ at = 0. Time is taken non-zero. ⇒ 'a' must be zero.
(Note: If you see answer in NCERT book, it gives an scenario, 'if the particle rebounds instantly with the same speed, it implies infinite acceleration which is unphysical a scenario' Can you explain this in the given content. Write it in comments section?)
(d) with positive value of acceleration must be speeding up. ✗ (False)
Consider a scenario when acceleration (a) is +ve and object is moving in opposite direction(-ve). E.g. a ball is thrown up (-ve direction), acceleration due to gravity (g) is acting downwards (+ve). Speed of the ball decreases.
The above statement can only be true, when both velocity and acceleration are in same direction.
Q12: 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.
Height from where ball is dropped (h) = 90 m
Initial velocity (u) = 0m
Acceleration (g) = 9.8 ms-2
Time taken by the ball to reach ground (t1) = ?
Final velocity when ball will strike the ground (v1) = ?
Using eqn. S = ut + ½(at2)
i.e. h = 0 + ½(gt12)
⇒ 90 = ½(gt12)
t12 = (90 ✕ 2)/9.8 = 18.378
⇒ t1 = √18.378 = 4.29s
Final velocity v1 = u + gt1 = 0 + (9.8)(4.29) = 42.04 m/s
Rebound velocity (u2) = (0.9)(42.04) = 37.84 m/s
After rebound, time taken (t'2) by the ball to reach maximum height is, when v2 = 0 m/s
-v2 = -u2 + gt'2
⇒ 0 + -u2 = gt'2
⇒ t'2 = u2 /g = 37.84/9.8 = 3.86s
At this stage, total time elapsed t2 = t'2 + t1 = 3.86 + 4.29 = 8.15s
⇒ At t2 = 8.15s, v2 = 0
Time taken by the ball to strike ground (t'3) = Time of descent (t'2)= 3.86s
And initial velocity before ball strikes the ground u3 = -v2 = 0
Velocity with which it will strike the ground second time v3 = u3+ gt'3 (or v3 = u2 )
v3 = 0 +9.8 ✕ 3.86 = 37.84 m/s
Next rebound velocity (u4 ) = (0.9) ✕ v3 = 0.9 ✕37.84 = 34.05 m/s
Total time taken = t2 + t'3 = 8.15 + 3.86 = 12.01 s
The velocity-time graph will look like:
Q3.13 Explain clearly, with examples, the distinction between :
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
Answer: Consider a student starts his trip from home to school (as shown). He covers the actual curved path (show with blue colour).
(b) As depicted, Actual Path (Distance) ≥ Magnitude of Displacement
⇒ Average Speed (= Distance/Time) ≥ Average Velocity (=Displacement/Time)
If the student moves along a straight line, Distance = magnitude of Displacement, in that case Average Speed = Magnitude of Average Velocity.
See page for Q14 onwards ...✈