Question

If the car’s speed decreases at a constant rate from 71 mi/h to 50 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:

The acceleration and the distance are 25200 mi/h² and 0.1008 mi.

Explanation:

Given that,

Initial speed = 71 mi/h

Final speed = 50 mi/h

Time = 3.0 s

(a). We need to calculate the acceleration

Using equation of motion

$v=u+at$

$a=\dfrac{v-u}{t}$

Put the value in the equation

$a=\dfrac{(50-71)\times3600}{3}$

$a=-25200\ mi/h^2$

Negative sign shows the deceleration.

(b). We need to calculate the distance

Using equation of motion

$v^2=u^2+2as$

$(50)^2=(71)^2+2\times(-25200)\times s$

$s=\dfrac{(50)^2-(71)^2}{-25200}$

$s=0.1008\ mi$

Hence, The acceleration and the distance are 25200 mi/h² and 0.1008 mi.