Question

A car braked with a constant deceleration of 18 ft/s2, producing skid marks measuring 100 ft before coming to a stop. How fast was the car traveling when the brakes were first applied?.

Answer 1

`60\; {\rm ft \cdot s^(-1)}`.

Step-by-step explanation:

Let
`a` denote the acceleration of this car;
`a = (-18)\; {\rm ft \cdot s^(-2)}`. Note that acceleration
`a\!` measures the rate of change in velocity. Since this car is slowing down, acceleration
`a\!\!` will be negative.

Let
`x` denote the displacement of the car. It is given that
`x = 100\; {\rm ft}` as the car braked.

Let
`v` denote the velocity of the car after braking. Since the car has stopped,
`v = 0\; {\rm ft \cdot s^(-1)}`.

Let
`u` denote the initial velocity of the car. Apply the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x` to find the value of
`u\!`. Rearrange this equation to obtain:

`u^(2) = v^(2) - 2\, a\, x`.

`\displaystyle u = \sqrt{v^(2) - 2\, a\, x}`.

Substitute in
`v = 0\; {\rm ft \cdot s^(-1)}`,
`a = 18\; {\rm ft \cdot s^(-2)}`, and
`x = 100\; {\rm ft}` to find the value of
`u`:

`\begin{aligned} u &= \sqrt{v^(2) - 2\, a\, x} \\ &= \left(\sqrt{0^(2) - 2 * 18 * 100}\right)\; {\rm ft \cdot s^(-1)} \\ &= 60\; {\rm ft \cdot s^(-1)}\end{aligned}`.

In other words, this car was travelling at
`60\; {\rm ft \cdot s^(-1)}` before braking.