Question

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 acceleration?

a) 1. 1.43 s, -1.75 m/s²
b) 2. 1.43 s, -1.40 m/s²
c) 3. 2.00 s, -1.75 m/s²
d) 4. 2.00 s, -1.40 m/s²

Answer 1

It takes the commuter 1.43 seconds to reach a speed of 2.00 m/s with an acceleration of 1.40 m/s², and when braking to a stop in 0.800 s, her deceleration is -2.50 m/s².

Step-by-step explanation:

To solve for how long it takes for the commuter to reach a speed of 2.00 m/s with an acceleration of 1.40 m/s², we can use the kinematic equation: v = u + at, where v is the final velocity, u is the initial velocity (0 m/s in this case, as she's starting from rest), a is the acceleration, and t is the time. Plugging in the values we get:
2.00 m/s = 0 m/s + (1.40 m/s²)(t). Solving for t gives us 2.00 m/s / 1.40 m/s² = 1.43 s.

For the deceleration when the commuter brakes to a stop in 0.800 s, we use the equation: a = Δv / Δt. Since the final velocity is 0 m/s (stopped) and the initial velocity before braking was 2.00 m/s, the change in velocity (Δv) is -2.00 m/s³ and the time (Δt) is 0.800 s. So, deceleration a = (-2.00 m/s) / (0.800 s) = -2.50 m/s².