The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.
a(t) = 2t + 4, v(0) = −5, 0 ≤ t ≤ 5
(a) Find the velocity at time t.
Find v(t) in m/s
Then, find the distance traveled during the given time interval.(in meters)
- cryptogramcornerLv 62 months ago
Acceleration is the rate of change of velocity, and so velocity is the antiderivative of acceleration. If a(t) = 2t+4 , then v(t) = t^2 + 4t + C. Knowing that v(0) = -5, we know that C is -5. v(t) = t^2 + 4t -5. . velocity is the rate of change of position, so position is the antiderivative of velocity. s(t) = t^3/3 +2t^2 -5t + C. To get the distance you just need to take s(5) - s(0) . All the terms except C in s(0) are zero, and the C in s(5) is cancelled out by the C in s(0), so the answer is just 5^3/3 + 2(5^2) - 5(5) = 125/3 + 50 - 25 = 41.66666 + 25 = 66.66666