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liberstina [14]
3 years ago
15

A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal

t?
Physics
1 answer:
Karolina [17]3 years ago
4 0

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2

Let d be the breaking distance. It can be calculated using third equation of motion as :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m

So, the required distance covered by the car is 44.64 m.

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Explanation:

Given that,

Mass of a freight car, m_1=30,000-kg

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Lost in kinetic energy is 8518.82. Negative sign shows loss.

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