Please answer the question below and show full working - How long does it take a 1500-Kg car to stop from a velocity of 45 m/s if a braking force of 4000 N used? If the braking force is increased 3-fold, would the car stop faster? How long would it take?
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Time to stop the car = 16.875 seconds
YES
5.625 seconds
Explanation:
Retardation = F / m
= 4000 ÷ 1500
= 2.6666666667 m/ s2
we know v = u + at
so t = ( v - u ) / a
= ( 0 - 45) ÷(- 2.6666666667)
= 16.875 seconds
If the braking force is increased 3-fold, would the car stop faster?
Yes
How long would it take?
5.625 seconds
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