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The driver of a car going 90.0 km/h suddenly sees the lights of a barrier 35.0 m ahead. It takes the driver 0.75 s to apply the...Asked by johnny
A driver of a car going 95.0 km/h suddenly sees the lights of a barrier 39.0 m ahead. It takes the driver 0.95 s before he applies the brakes, and the average acceleration during braking is -10.0 m/s2.
What is the maximum speed at which the car could be moving and not hit the barrier 39.0 m ahead? Assume the acceleration rate doesn't change.
What is the maximum speed at which the car could be moving and not hit the barrier 39.0 m ahead? Assume the acceleration rate doesn't change.
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Answered by
drwls
Ignore the 95.0 km/h speed of the driver. He may hit the barrier. They want you to calculate the speed V for which the stopping distance would be exactly 39.0 m.
The stopping distance will be
V*0.95 + (V/{a})*(V/2) = 39.0 m
Solve for V.
The second term, V^2/(2|a|)
is the distance travelled with brakes applied, It equals the average speed times the time spent braking. The first term, V/0.95s, it the distance travelled before the brakes are applied,
The stopping distance will be
V*0.95 + (V/{a})*(V/2) = 39.0 m
Solve for V.
The second term, V^2/(2|a|)
is the distance travelled with brakes applied, It equals the average speed times the time spent braking. The first term, V/0.95s, it the distance travelled before the brakes are applied,
Answered by
johnny
Can you go through and show all work to the finish?
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