Question

A car traveling at 17.2 m/s skids to a stop in 188 m from the point where the brakes were applied.

In what distance would the car have stopped had it been going 58.5 m/s , if tires and road condition were unchanged (i.e. if the acceleration were the same)?

Answer #1

Initial speed of the car = V_{1} = 17.2 m/s

Final speed of the car = V_{2} = 0 m/s (Comes to a
stop)

Distance the car travels before stopping = D_{1} = 188
m

Acceleration of the car = a

V_{2}^{2} = V_{1}^{2} +
2aD_{1}

(0)^{2} = (17.2)^{2} + 2a(188)

a = -0.7868 m/s^{2}

New initial speed of the car = V_{3} = 58.5 m/s

New final speed of the car = V_{4} = 0 m/s

New distance the car travels before stopping = D_{2}

All the conditions are same therefore the acceleration of the car is the same.

V_{4}^{2} = V_{3}^{2} +
2aD_{2}

(0)^{2} = (58.5)^{2} +
2(-0.7868)D_{2}

D_{2} = 2175 m

**Stopping distance of the car if it is going at a speed
of 58.5 m/s = 2175 m/s**

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