Question

uring the investigation of a traffic accident, police find skid marks 89.9 m long. They determine the coefficient of friction between the car's tires and the roadway to be 0.500 for the prevailing conditions. Estimate the speed of the car when the brakes were applied.

Answer 1

Answer:

$V_{0}=29.68m/s$

Explanation:

In order to solve this problem, we must first do a drawing of the situation (see attached picture).

When the brakes of the car were applied, we can see that there was only one horizontal force affecting the vehicle's movement, which was the force of friction. When analyzing the free body diagram we can apply Newton's laws to determine the equations we will use to solve this.

$\sum F_{x}=ma$

so:

-f=ma

we also know that:

$f=N\mu_{k}$

so

$-N\mu_{k}=ma$

we can now solve for the acceleration, so we get:

$a=\frac{-N\mu_{k}}{m}$

we don't know what the normal force is, so we can find it out by analyzing the vertical forces applied to the car:

$\sum F_{y}=0$

so we get:

N-W=0

which means that:

N=W

we also know that:

W=mg

so

N=mg

we can substitute this into the first equation so we get:

$a=\frac{-mg\mu_{k}}{m}$

which simplifies to:

$a=-g\mu_{k}$

we can now substitute the provided data:

$a=(-9.8m/s^{2})(0.5)$

which yields:

$a=-4.9m/s^{2}$

once we got the acceleration, we can use kinematics formulas to solve this, we got the following formula:

$a=\frac{V_{f}^{2}-V_{0}^{2}}{2x}$

we know the final velocity must be zero, since that's where the car got to a stop, so the formula then becomes:

$a=-\frac{V_{0}^{2}}{2x}$

we can now solve for the initial velocity, which yields:

$V_{0}=\sqrt{-2xa}$

so we can now substitute the daa we know, so we get:

$V_{0}=\sqrt{-2(89.9m)(-4.9m/s^{2})}$

so we get:

$V_{0}=29.68m/s$

So from this we know that the velocity of the car must have been of at least 29.68m/s when the brakes were applied.