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Q4 OQ

Expert-verifiedFound in: Page 137

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**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 **

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

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.

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

$\frac{1}{2}m{u}^{2}=F\xb7{d}_{1}......\left(1\right)$

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.

$\frac{1}{2}m{v}^{2}=F\xb7{d}_{2}\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{\left(\frac{u}{2}\right)}^{2}=F\xb7{d}_{2}\phantom{\rule{0ex}{0ex}}\frac{1}{4}\left(\frac{1}{2}m{u}^{2}\right)=F\xb7{d}_{2}......$

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

$\frac{\frac{1}{4}\left(\frac{1}{2}m{u}^{2}\right)}{\left(\frac{1}{2}m{u}^{2}\right)}=\frac{F\xb7{d}_{2}}{F\xb7{d}_{1}}\phantom{\rule{0ex}{0ex}}\frac{{d}_{2}}{d}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}{d}_{2}=\frac{d}{4}$

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.

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