A car starts from rest. It has to cover a distance of 500 m. the coefficient of friction between the road and
the tyre is 0.5. The minimum time in which the car can cover this distance is (g=10m/ s2 ).

Question

Aman Bansal · Answer

dear kaushik ram,

the car moves due to the frictional force provided due to the motion of tyres in opposite direction,

the frictional force = μmg

therefore acceleration provided due to friction is 0.5*10

and using second law of motion,

s=ut + 1/2*a*t²

t=200^1/2sec.

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A car starts from rest. It has to cover a distance of 500 m. the coefficient of friction between the road and the tyre is 0.5. The minimum time in which the car can cover this distance is (g=10m/ s2 ).

A car starts from rest. It has to cover a distance of 500 m. the coefficient of friction between the road and
the tyre is 0.5. The minimum time in which the car can cover this distance is (g=10m/ s2 ).

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A car starts from rest. It has to cover a distance of 500 m. the coefficient of friction between the road and the tyre is 0.5. The minimum time in which the car can cover this distance is (g=10m/ s2 ).

A car starts from rest. It has to cover a distance of 500 m. the coefficient of friction between the road and the tyre is 0.5. The minimum time in which the car can cover this distance is (g=10m/ s2 ).

1 Answers

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A car starts from rest. It has to cover a distance of 500 m. the coefficient of friction between the road and
the tyre is 0.5. The minimum time in which the car can cover this distance is (g=10m/ s2 ).