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3 years ago

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A 2.24x10^3 kg car slows down uniformly from 20.0 m/s to a stop (assume constant acceleration) via a braking force of 8.41x10^3

N . How far did the car travel during this time?

1 answer:

melisa1 [442]3 years ago

4 0

Answer:

Explanation: 139 m

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N=392

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$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

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$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

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