The velocity of a particle moving along a line is a function of time given by v(t) = 88 / (t^2 +9t +8). Find the distance that the particle has traveled after t=7 seconds if it started at t=0 seconds.

Answer 1

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Answer 2

v(t) = 88 / [(t+8)(t+1)] = A / (t+8) + B/(t+1)

88 = A(t+1) + B(t+8)

0t + 88 = (A+B)t + (A+8B)

So, A+B = 0 and A+8B = 88

We get B = 88/7 and A = -88/7

Distance = ∫_{(0 to 7)}v(t)dt = ∫_{(0 to 7)} [(-88/7)(1/(t+8)) + (88/7)(1/(t+1))]dt

= (88/7)[ln(t+1) - ln(t+8)]_{(0 to 7)}

=(88/7)[ln8 - ln15 - (ln1 - ln8)] = (88/7)[2ln8 - ln15] = (88/7)ln(64/15) ≈ 18.2