Question

A car travelling at 22.4 m/s skids to a stop in 2.55s. Determine the skidding distance of the car (assume uniform acceleration).

Answer 1

Approximately
`28.6\; {\rm m}`.

Step-by-step explanation:

Let
`u` denote the initial velocity of the vehicle, and let
`v` denote the velocity of the vehicle after skidding. It is given that the initial velocity was
`u = 22.4\; {\rm m\cdot s^(-1)}`. Since the vehicle skidded to a stop,
`v = 0\; {\rm m\cdot s^(-1)}`.

Let
`t` denote the duration of the skid. It is given that
`t = 2.55\; {\rm s}`.

Under the assumption that acceleration is constant, SUVAT equations will apply.

Specifically, the SUVAT equation
`x &= (1/2)\, (u + v)\, t` will be satisfied. In this equation, the displacement of the vehicle is equal to average velocity times duration. This equation allows the displacement
`x` to be found from
`u`,
`v`, and
`t` without knowing the exact value of acceleration:

`\begin{aligned}x &= \left((u + v)/(2)\right)\, t \\ &= \left((22.4 + 0)/(2)\; {\rm m\cdot s^(-1)}\right)\; (2.55\; {\rm s}) \\ &\approx 28.6\; {\rm m}\end{aligned}`.