Motion in Straight Line
NCERT Solution
Part 2
Question 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-1 in the same direction, with A ahead of B. The driver of B decides to overake 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?
Answer: Given, length of train = 400 m
V0 = 72 km h-1 `=72xx5/(18)=20` m s-1
Acceleration a = 1 m s-2 and t = 50 s
Relative velocity of train B with respect to A = 0
Distance can be calculated as follows:
`X=v_0t+1/2at^2`
`=0+1/2xx1xx50^2`
= 1250 m
Original distance = 1250 – 800 = 450 m
(Because total length of two trains = 800 m)
Question 8: On a two lane road, car A is travelling with a speed of 36 km h-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h-1 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?
Answer: Velocity of A = 36 km h-1 `=36xx5/(18)=10` m s-1
Velocity of B and C = 54 km h-1 `-54xx5/(18)=15` m s-1
Relative velocity of B with respect to A = 15 – 10 = 5 m s-1
Relative velocity of C with respect to A = 15 + 10 = 25 m s-1
Relative velocity of B with respect to C = 15 + 15 = 30 m s-1
AB = AC = 1 km = 1000 m
Time taken by C to overtake A `=(1000)/(25)=40` s
B needs to overtake A within 40 s to avoid accident
So, acceleration can be calculated as follows:
`X=v_0t+1/2at^2`
Or, `1000=5xx40+1/2a\xx40^2`
Or, 1000 = 200 + 800 a
Or, 800 a = 1000 – 200 = 800
Or, a = 1 m s-2
Question 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-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?
Answer: If vc is speed of cyclist and vB is speed of bus then
Relative speed of bus moving in same direction = vB - vc
Relative speed of bus moving in opposite direction = vB + vc
Bus moving in same direction overtakes the cyclist every 18 minutes. So, distance covered by bus in 18 min
`=(v_B-20)xx(18)/(60)`
Bus moving in opposite direction overtakes the cyclist every 6 minutes. So, distance covered by bus in 6 min
`=(v_B+20)xx(6)/(60)`
As the bus leaves after every T minutes so distance covered in by bus in T minute
`=v_B\xxT/(60)`
Now,
`(v_B-20)xx(18)/(60)=v_B\xxT/(60)` …………(1)
And
`(v_B+20)xx(6)/(60) =v_B\xxT/(60)` ……..(2)
Comparing above two equations, we get
`(v_B-20)xx(18)/(60)= (v_B+20)xx(6)/(60)`
Or, 3(vB - 20) = vB + 20
Or, 3vB = vB + 80
Or, 2vB = 80
Or, vB = 40 km h-1
Substituting this value in equation (1), we get:
`(40-20)xx(18)/(60)=40\xxT/(60)`
`T=(20xx18)/(40)=9` min
Question 10: A player throws a ball upwards with an initial speed of 29.4 m s-1
(a) What is the direction of acceleration during the upward motion of the ball?
Answer: Vertically downward
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
Answer: v = 0 and a = -9.8 m s-2
(c) Choose x = 0 m and t = 9 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.
Answer: During upward motion, position is positive, velocity is negative but acceleration is positive. During downward motion, position, velocity and acceleration are positive
(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)
Answer: v0 = 29.4 m/s, v = 0 m/s and a = -9/8 m/s2
Distance can be calculated as follows:
`v^2=v_0^2+2aX`
Or, 0 = (29.4)2 + 2 × (-9.8) × X
Or, 19.6 X = 864.36
Or, `X=(864.36)/(19.6)=44.1` m
Now, time can be calculated as follows:
v = u + at
Or, 0 = 29.4 – 9.8 t
Or, `t=(29.4)/(9.8)=3` s
Total time (ascent + descent) = 3 + 3 = 6 s
Question 11: Read each statement below carefully and state with reasons and examples, if it is true or false:
A particle in one-dimension motion
(a) With zero speed at an instance may have non-zero acceleration at that instant.
Answer: It is true. When an object is thrown upward its speed becomes zero at the highest point. At that instance its acceleration will still be equal to g.
(b) With zero speed may have non-zero velocity.
Answer: It is false, because a particle with non-zero velocity must have non-zero speed.
(c) With constant speed must have zero acceleration.
Answer: It is true
(d) With positive value of acceleration must be speeding up.
Answer: It is false, because it is possible only when direction of motion is same as direction of acceleration.