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27 November, 05:10

The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track and the train reaches it 76s later. The acceleration is - 0.26s^2. Find its speed at the next signal.

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  1. 27 November, 05:23
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    3.2 m/s

    Explanation:

    Given:

    Δx = 1000 m

    v₀ = 23 m/s

    a = - 0.26 m/s²

    t = 76 s

    Find: v

    This problem is over-defined. We only need 3 pieces of information, and we're given 4. There are several equations we can use. For example:

    v = at + v₀

    v = (-0.26 m/s²) (76 s) + (23 m/s)

    v = 3.2 m/s

    Or:

    Δx = ½ (v + v₀) t

    (1000 m) = ½ (v + 23 m/s) (76 s)

    v = 3.3 m/s

    Or:

    v² = v₀² + 2aΔx

    v² = (23 m/s) ² + 2 (-0.26 m/s²) (1000 m)

    v = 3.0 m/s

    Or:

    Δx = vt - ½ at²

    (1000 m) = v (76 s) - ½ (-0.26 m/s²) (76 s) ²

    v = 3.3 m/s

    As you can see, you get slightly different answers depending on which variables you use. Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.
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