Question

A 2000 kg car skids to a halt on a wet road where mk= 0.70. How fast was the car traveling if it leaves 75m long skid marks?

Answer 1

Given:

Mass of car, m = 2000 kg

Coefficient of kinetic friction, μk = 0.70

Distance = 75 m

Let's find the speed of the car.

To find the speed of the car, apply the motion formula:

`v^2=u^2+2as`

Where:

v is the final velocity = 0 m/s

u is the initial velocity

a is the acceleration of the car.

s is the distance = 75 m

To find the acceleration, we have:

`a=-\mu_k*g`

Where:

μk = 0.70

g is acceleration due to gravity = 9.8 m/s²

Thus, we have:

`\begin{gathered} a=-0.70*9.8 \\ \\ a=-6.86\text{ m/s}^2 \end{gathered}`

The deceleration of the car is -6.86 m/s².

Now, to find the initial velocity u, we have:

`\begin{gathered} v^2=u^2+2as \\ \\ 0^2=u^2+2(-6.86)(75) \\ \\ 0=u^2-1029 \\ \\ u^2=1029 \end{gathered}`

Take the square root of both sides:

`\begin{gathered} √(u^2)=√(1029) \\ \\ u=32.08\text{ m/s} \end{gathered}`

Therefore, the speed of the car was 32.08 m/s.

ANSWER:

32.08 m/s