Question

A train starting from rest accelerates uniformly for 100s, runs at a constant speed for 5min and then comes to a stop with uniform retardation in the next 150s. During this motion it covers a distance of 4.25km. Find its constant speed, acceleration and retardation.

Answer 1

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

$4.25\times10^{3}=\dfrac{1}{2}\times(450+200)\times v_{max}$

$v_{max}=\dfrac{2\times(4.25\times10^{3})}{650}$

$v_{max}=13.0\ m/s$

The maximum velocity is constant speed.

We need to calculate the acceleration

Using equation of motion

$v=u+at$

Put the value into the formula

$13.0=0+a\times100$

$a=\dfrac{13}{100}$

$a=0.13\ m/s^2$

The acceleration is 0.13 m/s².

We need to calculate the retardation

Using equation of motion

$v=u+at$

Put the value into the formula

$0=13+a\times150$

$a=\dfrac{-13}{150}$

$a=-0.086\ m/s^2$

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.

Answer 2

Explanation:

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