a 2.9x10^3 kg car accelerates from rest under the action of two forces. one is a

Question

a 2.9x10^3 kg car accelerates from rest under the action of two forces. one is a forward force of 1154 N provided by traction between the wheels and the road. the other is a 930 N resistive force due to various frictional forces. how far must the car travel for its speed to reach 2.5m/s? answer in units of m. a 2.9x10^3 kg car accelerates from rest under the action of two forces. one is a forward force of 1154 N provided by traction between the wheels and the road. the other is a 930 N resistive force due to various frictional forces. how far must the car travel for its speed to reach 2.5m/s? answer in units of m. a 2.9x10^3 kg car accelerates from rest under the action of two forces. one is a forward force of 1154 N provided by traction between the wheels and the road. the other is a 930 N resistive force due to various frictional forces. how far must the car travel for its speed to reach 2.5m/s? answer in units of m.

Explanation / Answer

The net force on the car will be:

1154 - 930 = ma

=> 224 = 2900a

=> a = 0.07724 m/s2

now use, S = [v2 - u2]/2a

=> S = [2.52 - 0]/(2 x 0.07724) = 40.457 meters. This is how far the car must travel for its speed to become 2.5 m/s.

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