The velocity of a particle moving along a line is a function of time given by v(t) = 72 / (t^2 + 12t + 35). Find the distance that the particle has traveled after t=8 seconds if it started at t=0 secs

Answer 1

The distance travelled x(t=8) - x(t=0) = ∫_t=0^t=8 v(t)dt, so all there is left to do is compute the antiderivative of v(t).

That is the ratio of polynomials, so we write 1/(t^2 + 12t + 35) = 1/[(t+5)(t+7)] = A/(t+5) + B/(t+7), and we need to determine the coefficients A and B.

Since A/(t+5) + B/(t+7) = ((A+B)t + (7A + 5B))/[(t+5)(t+7)] = 1/[(t+5)(t+7)], we have the systems:

A+B = 0 and 7A + 5B = 1 so that A = 1/2 and B = -1/2

So then: 1/(t^2 + 12t + 35) = 1/2 * 1/(t+5) -1/2 * 1/(t+7), so that ∫v(t)dt = 72/2 * (ln|x+5| - ln|x+7|).

Distance travelled = [72/2 * (ln|x+5| - ln|x+7|)]_t=0^t=8 = 36 (ln13 - ln15 - ln5 + ln7)