A commuter backs her car out of her garage with a constant acceleration of 1.40 m/s2.
(a) How long does it take her to reach a speed of 1.60 m/s?
s
(b) If she then brakes to a stop in 0.8 s, what is her (constant) deceleration?
m/s2
To solve these problems, we can use the equations of motion that relate distance, time, velocity, and acceleration.
For part (a):
We need to find the time it takes for the car to reach a speed of 1.60 m/s.
We can use the equation:
v = u + at
Where:
u = initial velocity (0 m/s as the car starts from rest)
v = final velocity (1.60 m/s)
a = acceleration (1.40 m/s^2)
t = time
Rearranging the equation to solve for time, we get:
t = (v - u) / a
Substituting the given values, we get:
t = (1.60 m/s - 0 m/s) / 1.40 m/s^2
t = 1.60 s / 1.40 m/s^2
t = 1.14 s
So, it takes her 1.14 seconds to reach a speed of 1.60 m/s.
For part (b):
We need to find the constant deceleration when the car brakes to a stop in 0.8 seconds.
We can use the equation:
v = u + at
Where:
u = initial velocity (1.60 m/s)
v = final velocity (0 m/s as the car stops)
a = acceleration (deceleration in this case)
t = time (0.8 s)
Rearranging the equation to solve for deceleration, we get:
a = (v - u) / t
Substituting the given values, we get:
a = (0 m/s - 1.60 m/s) / 0.8 s
a = -1.60 m/s / 0.8 s
a = -2 m/s^2
So, the constant deceleration of the car when braking is -2 m/s^2. The negative sign indicates deceleration.