Question

1. a commuter backs her car out of her garage with an acceleration of 1.40 m/s². (a) how long does it take her to reach a speed of 2.00 m/s? (b) if she then brakes to a stop in 0.800 s, what is her deceleration?

Explanation:

Step1: Use the acceleration - velocity formula for part (a)

We know the formula $v = v_0+at$. Here, $v_0 = 0$ (starts from rest), $a = 1.40\ m/s^{2}$, and $v = 2.00\ m/s$. We need to solve for $t$. Rearranging the formula gives $t=\frac{v - v_0}{a}$.
Substituting the values: $t=\frac{2.00 - 0}{1.40}=\frac{2.00}{1.40}\ s\approx1.43\ s$.

Step2: Use the acceleration - velocity formula for part (b)

The initial velocity for this part $v_0 = 2.00\ m/s$, the final velocity $v = 0$, and the time $t = 0.800\ s$. The acceleration formula is $a=\frac{v - v_0}{t}$.
Substituting the values: $a=\frac{0 - 2.00}{0.800}=\frac{- 2.00}{0.800}=-2.50\ m/s^{2}$. The negative sign indicates deceleration.

Answer:

(a) $1.43\ s$
(b) $- 2.50\ m/s^{2}$