Question

The driver of a speeding truck slams on the brakes and skids to a stop through a distance d. On another trial, the initial speed of the truck is half as large. What now will be the truck’s skidding distance? (a) $2\mathrm{d}$(b) $\sqrt{2}\mathrm{d}$ (c) d (d) d/2 (e) d/4

Short answer

The stopping distance of the truck for half the initial speed is $\frac{\text{d}}{\text{4}}$ , and Option (e) is correct.

Step-by-step solution

Identification of given data

The stopping distance for the first case is$d_1=d$ .

The initial speed in the second case is $v=\frac{u}{2}$

The work-energy theorem is used to find the stopping distance for half the initial speed in the second case.

Determine the stopping distance in the second case

Apply the work-energy theorem for the truck in the first case.

$\frac{1}{2}m{u}^2=F·d_1......\left(1\right)$

Here, m is the mass of the truck, u is the initial speed of the truck, and F is the braking force.

Apply the work-energy theorem for the truck in the second case.

$$\begin{gathered} \frac{1}{2}m{v}^2=F·d_2 \\ \frac{1}{2}m{\left(\frac{u}{2}\right)}^2=F·d_2 \\ \frac{1}{4}\left(\frac{1}{2}m{u}^2\right)=F·d_2...... \end{gathered}$$

Divide equation (2) by equation (1) to find the stopping distance of the truck in the second case.

$$\begin{gathered} \frac{\frac{1}{4}\left(\frac{1}{2}m{u}^2\right)}{\left(\frac{1}{2}m{u}^2\right)}=\frac{F·d_2}{F·d_1} \\ \frac{d_2}{d}=\frac{1}{4} \\ {d}_2=\frac{d}{4} \end{gathered}$$

The stopping distance of the truck for half the initial speed is $\frac{d}{4}$ .

Therefore, the stopping distance of the truck for half the initial speed is $\frac{d}{4}$, and Option (e) is correct.