The acceleration function (in m/s^2) and the initial velocity are given for a particle moving along a line.

Question

a(t) = t + 6, v(0) = 4, 0 ≤ t ≤ 11

(a) Find the velocity at time t.

(b) Find the distance traveled during the given time interval

Answer 1

Derivative of displacement (with respect to time) is velocity.
Derivative of velocity (with respect to time) is acceleration.
Thus,
v(t) = ∫ (a(t)) dt
v(t) = ∫ (t + 6) dt
v(t) = (1/2)t^2 + 6t + C

To find C, you know that at t = 0, v(0) = 4. You can substitute it there:
v(t) = (1/2)t^2 + 6t + C
v(0) = (1/2)(0)^2 + 6(0) + C = 4
C = 4

Thus, v(t) = (1/2)t^2 + 6t + 4

Then for the second question, to get D(t), integrate v(t) furthur and evaluate the resulting D(t) from t=0 to t=11.