Question

A train starting from rest accelerates uniformly for 100s, runs at a constant speed for 5 minutes &
then comes to rest with uniform retardation in next 150 seconds. During motion, it covers a distance
of 4.25 km. Find (i) the constant speed (ii) acceleration (iii) retardation​

Answer 1

Explanation:

Given that,

Distance = 4.25 km

Acceleration time = 100 s

Constant speed time = 5 min

Retardation time = 150 s

We know that,

In velocity time graph shows the distance.

The distance is area under the graph.

The maximum velocity is constant speed.

We need to calculate the maximum velocity

Using graph

Distance = Area under the graph

Put the value into the formula

The maximum velocity is constant speed.

We need to calculate the acceleration

Using equation of motion

Put the value into the formula

The acceleration is 0.13 m/s².

We need to calculate the retardation

Using equation of motion

Put the value into the formula

The retardation is -0.086 m/s².

Hence, The constant speed is 13.0 m/s.

The acceleration is 0.13 m/s².

The retardation is -0.086 m/s²