Two motorcycles travel along a straight road heading due north. At t = 0 motorcycle 1 is at x = 50.0 m and moves with a constant speed of 6.

Question

Two motorcycles travel along a straight road heading due north. At t = 0 motorcycle 1 is at x = 50.0 m and moves with a constant speed of 6.5 m/s; motorcycle 2 starts from rest at x = 0 and moves with constant acceleration. Motorcycle 2 passes motor cycle 1 at the time t = 10.0s. What is the speed of motorcycle 2 when it passes motorcycle 1?

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Khánh Gia 19 hours 2021-07-22T03:59:36+00:00 1 Answers 0 views 0

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    2021-07-22T04:00:43+00:00

    Answer:

    Vf = 23 m/s

    Explanation:

    First we need to find the distance covered by the motorcycle 2 when it passes motorcycle 1. Using the uniform speed equation for motorcycle 1:

    s₁ = v₁t₁

    where,

    s₁ = distance covered by motorcycle 1 = ?

    v₁ = speed of motorcycle 1 = 6.5 m/s

    t₁ = time = 10 s

    Therefore,

    s₁ = (6.5 m/s)(10 s)

    s₁ = 65 m

    Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,

    s₂ = s₁ + 50 m

    s₂ = 65 m + 50 m

    s₂ = 115 m

    Now, using second equation of motion for motorcycle 2:

    s₂ = Vi t + (1/2)at²

    where,

    Vi = initial velocity of motorcycle 2 = 0 m/s

    Therefore,

    115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²

    a = 230 m/100 s²

    a = 2.3 m/s²

    Now, using 1st equation of motion:

    Vf = Vi + at

    Vf = 0 m/s + (2.3 m/s²)(10 s)

    Vf = 23 m/s

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