Question

a commuter backs her car out of her garage with a constant acceleration of 1.30 m/s².
(a) how long in seconds does it take her to reach a speed of 2.40 m/s?
(b) if she then brakes to a stop in 0.8 s, what is her (constant) deceleration in m/s²?

Explanation:

Step1: Identify known values (part a)

Initial velocity $v_0=0\ \text{m/s}$, final velocity $v=2.40\ \text{m/s}$, acceleration $a=1.30\ \text{m/s}^2$

Step2: Use velocity formula for part a

Use $v = v_0 + at$, solve for $t$:
$t = \frac{v - v_0}{a}$
Substitute values: $t = \frac{2.40 - 0}{1.30}$

Step3: Calculate time for part a

$t = \frac{2.40}{1.30} \approx 1.85\ \text{s}$

Step4: Identify known values (part b)

Initial velocity $v_0=2.40\ \text{m/s}$, final velocity $v=0\ \text{m/s}$, time $t=0.8\ \text{s}$

Step5: Use velocity formula for part b

Use $v = v_0 + at$, solve for deceleration (magnitude of $a$):
$a = \frac{v - v_0}{t}$
Substitute values: $a = \frac{0 - 2.40}{0.8}$

Step6: Calculate deceleration for part b

$a = -3.00\ \text{m/s}^2$, magnitude is $3.00\ \text{m/s}^2$

Answer:

(a) $1.85$ s
(b) $3.00$ m/s²