Question

A train running along a straight track at 30m/s is slowed uniformly to a stop in 44s. Find the acceleration and the stopping distance.

Answer 1

Given data:

* The initial velocity of the train is 30 m/s.

* The time taken by the train is 44 s.

* The final velocity of the train is zero.

Solution:

(a). By the kinematics equation,

The acceleration of the train in terms of the change in velocity is,

`a=(v-u)/(t)`

where v is the final velocity, u is the initial velocity, and t is the time taken by the train to stop,

Substituting the known values,

`\begin{gathered} a=(0-30)/(44) \\ a=(-30)/(44) \\ a=-0.682ms^(-2) \end{gathered}`

Here, the negative sign indicates the decrease in the velocity with time,

Thus, the accerlation of the train is -0.682 meter per second squared.

(b). By the kinematics equation, the distance tarveled by the train is,

`v^2-u^2=2aS`

where S is the distance traveled by the train,

Substituting the known values,

`\begin{gathered} 0-30^2=2*(-0.682)* S \\ -900=-1.36* S \\ S=(900)/(1.36) \\ S=661.76\text{ m} \end{gathered}`

Thus, the stopping distance is 661.76 m.