Question

A train travels from one station to next. The driver of train A starts from rest at time, t = 0 and accelerates uniformly for first 20 seconds. At t = 20 seconds, train reaches its top speed 25 metre per second. Then travels with same speed for further 30 seconds. Before coming to rest, total time = 60 seconds. Another train B is travelling parallel to train A. It starts from rest at t = 0 and accelerates uniformly for first 10 seconds. At t = 10 seconds, it reaches its top speed = 30 metre/second. Then travel at the same speed for further 20 seconds. Before coming to rest, total time = 80 seconds. (i) Calculate the declaration of train A as it comes to rest. (ii) In which interval speed of train B is constant ? (iii) Initial speed of train A and B. (iv) What is the numerical of average velocity to average speed of an object when it is moving along a straight line ?​

Answer 1

Answer:

the answer of this question is 2.4 m s-2

Answer 2

Given:

Train A

• Case-1

Initial velocity (u1) of the train=0 m/s

Final Velocity (v1) of the train=25m/s

• Case-2

Initial velocity (u2) of the train=25m/s

Final velocity (v2) of the train=0m/s

Train B

• Case-1

Initial velocity (U1) of the train=0m/s

Final velocity (V1) of the train=30m/s

• Case-2

Initial Velocity (U2) of the train=30m/s

Final Velocity (V2) of the train=0m/s

To Find:

1. The deceleration of train A as it comes to rest.
2. The time interval during which the speed of train B is constant.
3. Initial speeds of trains A and B.
4. Magnitude (Numerical) of average velocity to the average speed of an object when it is moving along a straight line.

Solution:

Part-1 of the question-

To find the deceleration of train A as it comes to rest, we will consider Case-2 of Train A given above.

Let us represent the deceleration by 'a'.

According to the question,

Total time for Train-A=60s

The time during which it decelerates (t) =60-(20+30)

⇒60-50

⇒10s

∴Train A decelerates for the last 10 seconds.

Now, by using 'Third equation of motion,

v2-u2=at

⇒25-0=a×10

⇒25÷10=a

⇒a=2.5m/s²

∴Hence, the deceleration of train A as it comes to rest is 2.5m/s².

Part-2 of the question-

Now for train B, we will see the question carefully

The time during which train B accelerates uniformly=10s

The time during which train B has constant speed= Next 20s

⇒ from the 10th second to the 30th second.

∴The interval for which the speed of train B remains constant is from the 10th second to the 30th second of the total time interval.

Part-3 of the question-

Now, as the question mentions that both the trains A and B start from rest at t=0 time,

we know that

The initial speed of train A= Initial speed of train B= 0m/s

∴The initial speeds of both trains A and B are 0 m/s each.

Part-4 of the question-

As we know that, when a body travels in a straight line, then

Displacement = Distance

∴ Velocity = Speed

So, Average velocity= Average speed of a body traveling in a straight line.

Average Velocity÷ Average speed=1

Hence, the magnitude(numerical) of average velocity to an average speed of an object when it is moving along a straight line is 1.