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5 November, 13:38

A small object moves along the x-axis with acceleration ax (t) = - (0.0320m/s3) (15.0s-t). At t = 0 the object is at x = - 14.0 m and has velocity v0x = 8.90 m/s. What is the x-coordinate of the object when t = 10.0 s?

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  1. 5 November, 16:03
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    56.3 m

    Explanation:

    aₓ (t) = - 0.0320 m/s³ (15.0 s - t)

    Integrate to get velocity:

    ∫ dv = ∫ a dt

    vₓ (t) - v₀ₓ = ∫₀ᵗ aₓ (t) dt

    vₓ (t) - v₀ₓ = ∫₀ᵗ - 0.0320 m/s³ (15.0 s - t) dt

    vₓ (t) - v₀ₓ = - 0.0320 m/s³ (15.0 s t - ½ t²) |₀ᵗ

    vₓ (t) - 8.90 m/s = - 0.0320 m/s³ (15.0 s t - ½ t²)

    vₓ (t) = - 0.0320 m/s³ (15.0 s t - ½ t²) + 8.90 m/s

    vₓ (t) = - 0.480 m/s² t + 0.0160 m/s³ t² + 8.90 m/s

    Integrate again to get position:

    ∫ dx = ∫ v dt

    x (t) - x₀ = ∫₀ᵗ vₓ (t) dt

    x (t) - x₀ = ∫₀ᵗ (-0.480 m/s² t + 0.0160 m/s³ t² + 8.90 m/s) dt

    x (t) - x₀ = (-0.240 m/s² t² + 0.00533 m/s³ t³ + 8.90 m/s t) |₀ᵗ

    x (t) - (-14.0 m) = - 0.240 m/s² t² + 0.00533 m/s³ t³ + 8.90 m/s t

    x (t) = - 0.240 m/s² t² + 0.00533 m/s³ t³ + 8.90 m/s t - 14.0 m

    Evaluate at t = 10 s:

    x (10) = - 0.240 m/s² (10 s) ² + 0.00533 m/s³ (10 s) ³ + 8.90 m/s (10 s) - 14.0 m

    x (10) = - 24.0 m + 5.33 m + 89.0 m - 14.0 m

    x (10) = 56.3 m
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