Question

A car moving on a horizontal road starts to decelerate at t-Os from an unknown initial velocity v. It slows down to 18m/s at te 10s and stops at t-16 s. If the acceleration of the car is constant during the whole trip (from Os to 1os), then the initial velocity v, is equal to:

Answer 1

Given

v_i = ?

v_m = 18 m/s

v_f = 0 m/s

Procedure

The first thing we are going to do is to calculate the acceleration from second 10 to second 16.

`\begin{gathered} v_f=v_o+at \\ 0m/s=18m/s+a*\lparen16-10s) \\ a=(-18m/s)/(6) \\ a=-3m/s^2 \end{gathered}`

We can now calculate the value of the initial velocity from the complete trip data

`\begin{gathered} v_f=v_o+at \\ 0=v_o-3m/s^2*16s \\ v_o=48m/s \end{gathered}`

The initial velocity would be 48 m/s