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spayn [35]
3 years ago
9

A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping dista

nce of the car?
Physics
1 answer:
qaws [65]3 years ago
5 0

<em>v</em>² - <em>u</em>² = 2 <em>a</em> ∆<em>x</em>

where <em>u</em> = initial velocity (27 m/s), <em>v</em> = final velocity (0), <em>a</em> = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆<em>x</em> = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆<em>x</em>

∆<em>x</em> = (27 m/s)² / (16 m/s²)

∆<em>x</em> ≈ 45.6 m

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Answer:

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