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White raven [17]

3 years ago

12

HELP!!!!!!!!!!!!

1 answer:

ziro4ka [17]3 years ago

4 0

Answer:

-4,200,000 J

Explanation:

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3 years ago

A car traveling at 91.0 km/h approaches the turn off for a restaurant 30.0 m ahead. If the driver slams on the brakes with the a

eduard

Answer: 49.92 m

Explanation:

In this situation the following equation will be useful:

$V^{2}=V_{o}^{2} +2 a d$

Where:

$V=0 m/s$ is the final velocity of the car, when it finally stops

$V_{o}=91 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=25.27 m/s$ is the initial velocity of the car

$a=-6.4 m/s^{2}$ is the constant acceleration of the car after the driver slams on the brakes

$d$ is the stopping distance

Isolating $d$ :

$d=\frac{-V_{o}^{2}}{2a}$

$d=\frac{-(25.27 m/s)^{2}}{2(-6.4 m/s^{2})}$

$d=41.919 m \approx 41.92 m$

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4 years ago