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19 January, 13:19

If a 1300-kg car can accelerate from 35 km? h to 65 km? h in 3.8 s, how long will it take to accelerate from 55 km? h to 95 km? h? Assume the power stays the same, and neglect frictional losses.

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  1. 19 January, 14:57
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    6.76 s

    Explanation:

    Power = Energy/time

    P = E/t ... Equation 1

    Where P = Power, E = kinetic energy, t = time.

    But,

    E = 1/2m (Δv) ² ... Equation 2

    Where m = mass of the car, Δv = change in velocity.

    Substitute equation 2 into equation 1

    P = 1/2m (Δv) ²/t ... Equation 3

    Given: m = 1300 kg, Δv = 65-35 = 30 km/h = 30 (1000/3600) m/s = 8.33 m/s, t = 3.8 s

    Substitute into equation 3

    P = 1/2 (1300) (8.33²) / 3.8

    P = 11869.15 W.

    Assuming the power stays the same,

    making t the subject of formula in equation 3

    t = 1/2m (Δv) ²/P ... Equation 4

    Given: m = 1300 kg, Δv = 95-55 = 40 km/h = 11.11 m/s, P = 11869.15 W

    Substitute into equation 4

    t = 1/2 (1300) (11.11²) / 11869.15

    t = 6.76 s
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