A car is traveling at a constant negative velocity of 12.0 m/s. The driver steps on the brakes. The car slows to a stop. The car travels 10.0 m while it is slowing down. What is the magnitude of its acceleration? [HINT: Pay attention to direction.]
________m/s2
To find the magnitude of the car's acceleration, we need to calculate the change in velocity and divide it by the time it takes to stop.
First, let's determine the time it takes for the car to stop. Since the car is already traveling at a constant velocity of -12.0 m/s, it slows down until it comes to a stop, so its final velocity is 0 m/s.
We can use the equation of motion:
vf = vi + at
Where:
vf is the final velocity (0 m/s),
vi is the initial velocity (-12.0 m/s),
a is the acceleration, and
t is the time.
Rearranging the equation, we get:
t = (vf - vi) / a
Since vf is 0 m/s and vi is -12.0 m/s, we can substitute these values:
t = (0 - (-12.0)) / a
t = 12.0 / a
Now, let's calculate the distance traveled while the car is slowing down. We are given that the car travels 10.0 m during this time.
Using the kinematic equation:
d = vit + (1/2)at^2
Where:
d is the distance (10.0 m),
vi is the initial velocity (-12.0 m/s),
a is the acceleration, and
t is the time.
Since we have already calculated t as 12.0 / a, we can substitute this value:
10.0 = -12.0 * (12.0 / a) + (1/2) * a * (12.0 / a)^2
10.0 = -144.0 / a + 144.0
10.0 - 144.0 = -144.0 / a
-134.0 = -144.0 / a
Now, let's solve for a:
a = -144.0 / -134.0
Simplifying, we get:
a = 1.075 m/s^2
Therefore, the magnitude of the car's acceleration is 1.075 m/s^2.