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,
[tex]a=\frac{v-u}{t}[/tex]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,
[tex]\begin{gathered} a=\frac{0-30}{44} \\ a=\frac{-30}{44} \\ a=-0.682ms^{-2} \end{gathered}[/tex]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,
[tex]v^2-u^2=2aS[/tex]where S is the distance traveled by the train,
Substituting the known values,
[tex]\begin{gathered} 0-30^2=2\times(-0.682)\times S \\ -900=-1.36\times S \\ S=\frac{900}{1.36} \\ S=661.76\text{ m} \end{gathered}[/tex]Thus, the stopping distance is 661.76 m.
