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?
To answer part (a) of the question, we need to use the equations of motion. The equation that relates acceleration, initial velocity, final velocity, and time is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity (u) is 0 m/s because the car starts from rest. The acceleration (a) is given as 1.40 m/s^2, and the final velocity (v) is 2.00 m/s. We need to find the time (t).
Rearranging the equation, we have:
t = (v - u) / a
Plugging in the values, we find:
t = (2.00 m/s - 0 m/s) / 1.40 m/s^2
Simplifying further, we have:
t = 1.43 s
Therefore, it takes the commuter approximately 1.43 seconds to reach the speed of 2.00 m/s.
Moving on to part (b) of the question, we need to find the deceleration (a) of the car. Deceleration is just the negative of the acceleration, so deceleration (a) = -1.40 m/s^2.
Given that the time (t) taken to stop is 0.800 s, we can use the equation:
v = u + at
Here, the final velocity (v) is 0 m/s since the car comes to a stop, and the initial velocity (u) is 2.00 m/s. We need to find the deceleration (a).
Rearranging the equation, we get:
a = (v - u) / t
Plugging in the values, we find:
a = (0 m/s - 2.00 m/s) / 0.800 s
Simplifying further, we have:
a = -2.50 m/s^2
Therefore, the deceleration of the car is approximately -2.50 m/s^2. The negative sign indicates deceleration or slowing down.