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.
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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?
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What is that?? Please tell us
Answer:
- Our final answer is 15 kg .
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Answer:
m/s^2
Explanation:
Force = mass × acceleration
kgm/s^2 = kg × acceleration
where acceleration = Force ÷ mass
= kg m/s^2 ÷ kg
:Acceleration = m/s^2
Answer:c
Explanation: because if you work it in a paper it should like lil wit is straight the numbers are going up by 16
