A commuter backs her car out of her garage with a constant acceleration of 1.20 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?
Are you Daniel Spry?
To solve this problem, we can use the equations of motion that relate distance, velocity, acceleration, and time. Here are the steps to solve the problem:
(a) To find how long it takes to reach a speed of 1.60 m/s, we need to find the time it takes to accelerate from 0 m/s to 1.60 m/s.
We can use the equation:
v = u + at
Where:
v = final velocity = 1.60 m/s
u = initial velocity = 0 m/s
a = acceleration = 1.20 m/s^2
t = time
Rearranging the equation to solve for time, we get:
t = (v - u) / a
Plugging in the values, we have:
t = (1.60 m/s - 0 m/s) / 1.20 m/s^2
t = 1.60 m/s / 1.20 m/s^2
t = 1.33 s
So, it takes her 1.33 seconds to reach a speed of 1.60 m/s.
(b) To find the deceleration when she brakes to a stop in 0.8 s, we need to find the magnitude of the acceleration.
We can use the equation:
a = (v - u) / t
Where:
a = acceleration
v = final velocity = 0 m/s
u = initial velocity = 1.60 m/s
t = time = 0.8 s
Plugging in the values, we have:
a = (0 m/s - 1.60 m/s) / 0.8 s
a = -1.60 m/s / 0.8 s
a = -2.0 m/s^2
The negative sign indicates that the car is decelerating or braking. Therefore, the constant deceleration is 2.0 m/s^2.
v = a t
1.6 = 1.2 t
t = 1.33 seconds
Vi = 1.60
v = Vi + a t
0 = 1.60 + a * 0.8
a = - 2 m/s^2