The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.

Question

a(t) = 2t + 2, v(0) = −3, 0 ≤ t ≤ 4

(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) = ∫ (2t + 2) dt
v(t) = t^2 + 2t + C

It was said that at t = 0, v(0) = -3.
v(0) = (0)^2 + 2(0) + C = -3
v(0) = C = -3

Substitute,
v(t) = t^2 + 2t - 3

For the second question, evaluate v(t) at the interval from t=0 to t=4. Hope this helps~ `u`

Answer 2

I mean for the second question, integrate v(t) further and evaluate from t=0 to t=4:
D(t) = ∫ v(t)
D(t) = ∫ (t^2 + 2t - 3)
D(t) = (1/3)t^3 + t^2 - 3t + C

at t = 0:
D(0) = (1/3)(0)^3 + (0)^2 - 3(0) + C
D(0) = C

at t = 4:
D(4) = (1/3)(4)^3 + (4)^2 - 3(4) + C
D(4) = 76/3 + C

Thus,
76/3 + C - C = 76/3 meters