A car is traveling at 28 m/s when the driver spots a large pothole in the road a distance 37 m ahead. She immediately applies her brakes. If her acceleration is -7.8 m/s2, does she manage to stop before reaching the pothole? What is this stopping distance?
The time required to stop is
T = Vo/|a| = 28/7.8 = 3.59 seconds.
The average speed during that time is Vo/2 = 14 m/s. The distance travelled while decelerating is
Vav*T = (Vo/2)*T = 14 m/s*3.59s = 50.26 m
She cannot stop in time to miss the pothole.
To determine if the car manages to stop before reaching the pothole, we can calculate the stopping distance using the given information.
First, we need to find the time it takes for the car to stop. We can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s because the car comes to a stop)
u = initial velocity (28 m/s)
a = acceleration (-7.8 m/s^2)
s = stopping distance (what we are trying to calculate)
Plugging in the values into the equation, we have:
0^2 = 28^2 + 2(-7.8)s
0 = 784 - 15.6s
Rearranging the equation to solve for s, we get:
15.6s = 784
s = 784 / 15.6
s ≈ 50.26 m
The stopping distance is approximately 50.26 m.
Since the stopping distance is greater than the distance to the pothole (37 m), the car will not manage to stop before reaching the pothole.