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A 2.9 × 103 kg car accelerates from rest under the action of two forces. One is a forward force of 1148 N provided by traction between the wheels and the road. The other is a 941 N resistive force due to various frictional forces. How far must the car travel for its speed to reach 2.7 m/s? Answer in units of m.

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Answer:

The car must travel 51.34 m for its speed to reach 2.7 m/s

Explanation:

Mass of car = 2.9 x 10³ kg

Forward force = 1148 N

Resistive force = 941 N

Total force = 1148 - 941 = 207 N

We know

            Force = Mass x Acceleration

              207 = 2.9 x 10³ x Acceleration

            Acceleration = 0.071 m/s²  

Now we have equation of motion, v² = u² + 2as

      Initial velocity, u = 0 m/s

      Final velocity, v = 2.7 m/s

      Acceleration, a = 0.071 m/s²  

Substituting

        v² = u² + 2as

       2.7² = 0² + 2 x 0.071 x s

        s = 51.34 m

The car must travel 51.34 m for its speed to reach 2.7 m/s