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Honors Physics

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An automobile accelerates from rest at
1.3 m/s
for 19 s. The speed is then held
constant for 23 s, after which there is an acceleration of −0.9 m/s
until the automobile
What total distance was traveled?
Answer in units of km

  • Honors Physics - ,

    while accelerating:
    distance=1/2 1.3 19^2

    constant speed period.
    distance= finalvelocityabove*23 seconds

    deaccelerating period.
    Vf^2=Vi^2 + 2ad where Vf=0 and Vi equals the finalvelocityabove. Solve for distance.

    time deaccelerating: 0=Vi+at solve for t

    add the total distances, and if you need, add the total times.

  • Honors Physics - ,

    I am still confused. For finalvelocity, where did you get 18?

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