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Degger [83]
2 years ago
9

A car moving 18.2 m/s starts to coast up a frictionless 10.0 degree hill. how far does it roll before coming to a stop

Physics
1 answer:
Kazeer [188]2 years ago
8 0

The distance covered by the car before it stops is 98.58 m.

<h3>Calculation:</h3>

Force acting on the car will be

ma = -mg sinθ

where,

m = mass of the car

a = acceleration of the car

g = acceleration due to gravity

θ = angle of inclination

Given,

θ = 10°

Put the values in the above equation,

ma = -mg sinθ

a = -g sin 10°

a = -10 (0.173)

a = -1.73 m/s²

We know the initial velocity is 18.2m/s and when the car comes to a stop the final velocity becomes zero.

u = 18.2 m/s

v= 0 m/s

Using one of the kinematics equations,

v² - u² = 2as

0 - (18.2)² = 2 (-1.73) (s)

331.24 = 3.36 (s)

s = 98.58 m

Hence, the distance covered by the car before it stops is 98.58 m.

Learn more about the other way to solve such numerical here:

brainly.com/question/12985617

#SPJ4

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