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?

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