You are driving through town at 14 m/s when suddenly a car backs out of a driveway in front of you. You apply the brakes and begin decelerating at 3.3 m/s2.How much time does it take to stop? After braking half the time found in part A, is your speed 7.0 m/s, greater than 7.0 m/s, or less than 7.0 m/s?
Vi = 14
a = -3.3
v = Vi + a t
at stop, v = 0
0 = 14 - 3.3 t
3.3 t = 14
for half time
if t = 7/3.3
v = 14 - 3.3 (7/3.3)
v = 14 - 7 = 7 on the nose
then
6s
To find the time it takes to stop, we can use the kinematic equation:
v = u + at
Where:
v = final velocity (0 m/s, since you're stopping)
u = initial velocity (14 m/s)
a = acceleration (-3.3 m/s^2, as it's deceleration in this case)
t = time
Rearranging the equation to solve for time t:
t = (v - u) / a
Plugging in the values:
t = (0 - 14) / (-3.3)
t = 14 / 3.3
t ≈ 4.24 seconds
So, it takes approximately 4.24 seconds to stop.
Now, to find the speed after braking half the time found in part A (which is half of 4.24 seconds), we can simply subtract the distance covered during that time from the initial velocity.
The distance covered during half the time is given by:
s = ut + (1/2)at^2
Since we know that s = 7.0 m (half the distance covered when stopping), and t = 4.24 seconds / 2 = 2.12 seconds, we can rearrange the equation to solve for u (final velocity after half the time):
u = (2s - at^2) / (2t)
Plugging in the values:
u = (2 * 7.0 - (-3.3) * (2.12)^2) / (2 * 2.12)
u ≈ 11.017 m/s
Therefore, the speed after braking half the time is approximately 11.017 m/s, which is greater than 7.0 m/s.