Question

a woman backs her car out of her garage with an acceleration of 2.80m/s^2. how long does it take her to reach a speed of 4.00m/s?

Answer 1

We can find the time that woman takes to reach a speed of 4.00 m/s from the kinematic equation:

$$v = v_0 + at,$$

here, $v_0 = 0 \ \dfrac{m}{s}$ is the initial velocity of the car, $v = 4.0 \ \dfrac{m}{s}$ is the final velocity of the car, $a = 2.80 \ \dfrac{m}{s^2}$ is the acceleration of the car and $t$ is the time that woman takes to reach a speed of 4.00 m/s.

Then, we get:

$$t = \dfrac{v-v_0}{a} = \dfrac{4.0 \ \dfrac{m}{s}-0 \ \dfrac{m}{s}}{2.80 \ \dfrac{m}{s^2}} = 1.43 \ s.$$

Answer:

$t = 1.43 \ s.$