A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 19

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The car will take 300 m before it stops due to applying break.

What's the relation between initial velocity, final velocity, acceleration and distance?

  • As per Newton's equation of motion, - = 2aS
  • V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
  • Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
  • So, 0² - 60² = 2×6× S

=> -3600 = -12S

=> S = 3600/12 = 300 m

Thus, we can conclude that the distance covered by the car is 300 m before it stopped.

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Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?

Learn more about the Newton's equation of motion here:

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