Physics

Two cars approach each other from an initial distance of 2400 m. Car A is moving to the E at a constant rate speed of 40ms. Car B is moving to the W at a speed of 10ms but is accelarating at a rate of 2.00ms^2.

What is the time before the cars meet

the distance a traveled before meeting b

the relative speed at which the cars hit.

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  1. Call x = 0 the location were Car A starts. Call that car's location X1.

    Car B then starts at x = 2400. Call its location X2

    X1 = 40 t

    X2 = 2400 - 10 t - t^2.

    They meet when X1 = X2.

    40t = 2400 -10 t - t^2

    Rearrange that to read
    t^2 +50 t -2400 = 0
    (t +80)(t-30) = 0

    Solve for the time t that is a positive root of that equation.

    Use that value of t to solve for the location where they meet. The relative velocity is V1 - V2. But V2 is negative, so you end up adding the speeds of the cars at time t.

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    posted by drwls

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