QuestionThis question is typical on some driver's license exams: A car moving at $42 \mathrm{~km} / \mathrm{h}$ skids 12 m with locked brakes.
How far will the car skid with locked brakes at $126 \mathrm{~km} / \mathrm{h}$ ? Assume that energy loss is due only to sliding friction.
Answer in units of $m$ .

Studdy Solution

STEP 1

1. The car's kinetic energy is entirely converted into work done against friction.
2. The frictional force is constant.
3. The relationship between speed and skid distance is quadratic because kinetic energy is proportional to the square of the speed.

STEP 2

1. Understand the relationship between speed, kinetic energy, and skid distance.
2. Set up the equation relating initial speed and skid distance.
3. Calculate the new skid distance using the relationship.

STEP 3

The kinetic energy of a car is given by $KE = \frac{1}{2}mv^2$ , where $m$ is mass and $v$ is velocity.
The work done by friction is $W = f \cdot d$ , where $f$ is the frictional force and $d$ is the skid distance.
Since all kinetic energy is converted into work done against friction, we have: $\frac{1}{2}mv^2 = f \cdot d$

STEP 4

For the initial speed $v_1 = 42 \, \text{km/h}$ and skid distance $d_1 = 12 \, \text{m}$ , we have: $\frac{1}{2}m(42)^2 = f \cdot 12$
For the new speed $v_2 = 126 \, \text{km/h}$ , the equation becomes: $\frac{1}{2}m(126)^2 = f \cdot d_2$

STEP 5

Divide the second equation by the first to eliminate $m$ and $f$ : $\frac{(126)^2}{(42)^2} = \frac{d_2}{12}$
Simplify the left side: $\left(\frac{126}{42}\right)^2 = \frac{d_2}{12}$
$\left(3\right)^2 = \frac{d_2}{12}$
$9 = \frac{d_2}{12}$
Solve for $d_2$ : $d_2 = 9 \times 12$
$d_2 = 108$
The car will skid:
$\boxed{108 \, \text{m}}$

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