Answer:
x = 54.3m (on the +ve x axis)
Explanation:
This is an Initial Value Problem. That means the initial values of certain parameters have been given and that can help solve the problem.
Given that acceleration, a, is:
ax(t) = - 0.032(15.0 - t)
And the initial values are:
x(t = 0) = - 14.0m
v(t = 0) = 8.7m/s
Hence,
a = - 0.032(15 - t)
a = - 0.48 + 0.032t
a = dv/dt = -0.48 + 0.032t
To obtain the velocity, v, integrate the acceleration and apply the initial values of v and t:
v = ∫dv/dt = ∫(-0.48 + 0.032t)
∫dv = ∫(-0.48 + 0.032t)dt
(v - v₀) = -0.48(t - t₀) + 0.032(t²/2 - t₀²/2)
Inputting the initial values t₀ = 0s, v₀ = 8.7m/s:
=> v - 8.7 = -0.48t + 0.032t²/2
v = 8.7 - 0.48t + 0.016t²
To obtain distance, x, integrate the velocity and apply the initial values:
v = dx/dt = 8.7 - 0.48t + 0.016t²
=> ∫dx/dt = ∫(8.7 - 0.48t + 0.016t²)
∫dx= ∫(8.7 - 0.48t + 0.016t²)dt
(x - x₀) = 8.7(t - t₀) - 0.48(t²/2 - t₀²/2) + 0.016(t³/3 - t₀³/3)
Inputting the initial values t₀ = 0s, x₀ = - 14.0m:
(x + 14.0) = 8.7t - 0.48t²/2 + 0.016t³/3
x = 8.7t - 0.48t²/2 + 0.016t³/3 - 14.0
Now that the distance, x, has been obtained, when t = 10s:
x = 8.7*10 - 0.48*10²/2 + 0.016*10³/3 - 14.0
x = 87 - 24 + 5.3 - 14.0
x = 54.3m
Therefore, at time, t = 10s, x = +54.3m. (i.e. 54.3 on the +ve x axis).
