Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a ramp at a high speed v0 and travels 690 ft in 8.5 s at constant deceleration before its speed is reduced to v0/2. Assume the same constant deceleration. a) Determine the additional time required for the truck to stop. The additional time required for the truck to stop is s b) the additional distance traveled by the truck is ft

a) The additional time required for the truck to stop is 8.5 seconds

b) The additional distance traveled by the truck is 230.05 ft

Explanation:

Since the acceleration is constant, the average speed is:

(final speed - initial speed) / 2 = 0.75 v0

Since travelling at this speed for 8.5 seconds causes the vehicle to travel 690 ft, we can solve for v0:

0.75v0 * 8.5 = 690

v0 = 108.24 ft/s

The speed after 8.5 seconds is: 108.24 / 2 = 54.12 ft/s

We can now use the following equation to solve for acceleration:

a = -6.367 m/s^2

Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds

Total distance traveled:

0 - 108.24^2 = 2 * (-6.367) * s

solving for s we get total distance traveled = 920.05 ft

Additional Distance Traveled: 920.05 - 690 = 230.05 ft

answer:

the answer is d

explanation: