Question

Position-time equation for the train is `x_t=2.6m+(8.1m/s)t+(2.6m/s^2)t^2`. What is (A) its initial velocity (B) its acceleration?

Answer 1

Part A - Initial velocity is `8.1m/s`
Part B - Accelaration is `5.2 m/s^2`

Explanation:

As position-time equation for the train is `x_t=2.6m+(8.1m/s)t+(2.6m/s^2)t^2`

velocity is given by `v_t=(dx_t)/dt=8.1m/s+5.2tm/s`

and initial velocity is `v_t(t=0)=8.1m/s+5.2xx0m/s=8.1m/s`

and accelaration is `a_t=(dv_t)/(dt)=(d^2x_t)/(dt^2)=5.2m/s^2`

As it is independent of `t`, it is uniform accelaration of `5.2m/s^2`