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23 October, 02:14

A bike first accelerates from 0.0m/s to 4.5m/s in 4.5 s, the continues at this constant speed for another 6.0 s. What is the total distance traveled by the bike?

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  1. 23 October, 02:45
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    37.125 m

    Explanation:

    Using the equation of motion

    s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration

    Distance during acceleration

    Acceleration, a=/frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.

    Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be

    a=/frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}

    Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion

    s=0*4.5 + 0.5*1*4.5^{2}=0+10.125 = 10.125 m

    Distance at a constant speed

    At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time

    Distance=4.5 m/s*6 s=27 m

    Total distance

    Total=27+10.125=37.125 m
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