You are driving a car 33 down the road. The stoplight in front of you suddenly turns red and you slam on the brakes, skidding to a halt. You get out of the car and measure the skid marks made by your tires to be 23.5 meters (Assume positive x to be in the forward direction)
What is the acceleration of the car?
If your car were going twice the velocity, how far would it take to stop if the acceleration were the same as for part (B)?
How long would it take the car to stop if it were going at twice the velocity?
vf^2=vi^2+2ad solve for a.
wouldn't it take 4 times longer?
32
To find the acceleration of the car, we can use the kinematic equation:
v^2 = u^2 + 2as
Here:
v = final velocity (0 m/s since the car comes to a stop)
u = initial velocity (initial speed of the car = 33 m/s)
a = acceleration (what we need to find)
s = distance covered (skid marks = 23.5 m)
To find the acceleration, we rearrange the equation:
a = (v^2 - u^2) / (2s)
Substituting the values:
a = (0^2 - 33^2) / (2 * 23.5)
a = (-1089) / 47
a = -23.21 m/s^2
So, the acceleration of the car is approximately -23.21 m/s^2 (negative sign indicates deceleration).
If the car were going twice the velocity (66 m/s) and the acceleration remained the same (-23.21 m/s^2), we can use the same equation to find the distance covered (s):
v^2 = u^2 + 2as
Rearranging the equation to solve for the distance (s):
s = (v^2 - u^2) / (2a)
Substituting the values:
s = (0^2 - 66^2) / (2 * -23.21)
s = (-4356) / (-46.42)
s = 93.76 m
So, if the car were going twice the velocity, it would take approximately 93.76 meters to stop, assuming the acceleration remained the same.
To find the time it would take for the car to stop if it were going twice the velocity (66 m/s), we can use the kinematic equation:
v = u + at
Rearranging the equation to solve for time (t):
t = (v - u) / a
Substituting the values:
t = (0 - 66) / -23.21
t = 2.84 seconds
So, if the car were going at twice the velocity, it would take approximately 2.84 seconds to stop, assuming the acceleration remained the same.