Question

If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, assuming that it continues to move in a straight line? What distance does the car travel during the braking period?

Answer 1

Answer:$3.874 m/s^2$

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

$60mi/h \approx 26.8224m/s$

$34mi/h \approx 15.1994 m/s$

we know acceleration is given by $=\frac{velocity}{Time}$

$a=\frac{15.1994-26.8224}{3}$

$a=-3.874 m/s^2$

negative indicates that it is stopping the car

Distance traveled

$v^2-u^2=2as$

$\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s$

$s=\frac{488.419}{2\times 3.874}$

s=63.038 m