Answer:
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D= vt +.5at^2
since he started at rest, v (initial velocity) is 0
so d=.5at^2
d = .5 (6m/s^2) (4.1s)^2
then put that into a calculator.
The car will take 300 m before it stops due to applying break.
<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
- As per Newton's equation of motion, V² - U² = 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.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
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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Answer:
Check the attached image
Explanation:
To solve the problem for time you will have to use the formula for time, t = d/s which means time equals distance divided by speed.
Kindly check the attached image below for the step by step explanation to the question.
