# Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 58

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A particle moves along a line so that its acceleration of time $t$ is $a(t) = 2t + 3$ $\left( \displaystyle \text{ measured in }\frac{m}{s^2} \right)$ with the initial velocity $v(0) = -4$.

a.) Find the velocity at time $t$

\begin{aligned} \int a(t) dt &= \int (2t + 3) dt\\ \\ \int a(t) dt &= 2 \int t dt + \int 3 dt\\ \\ \int a(t) dt &= 2 \left( \frac{t^{1+1}}{1+1} \right) + 3 \left( \frac{t^{0+1}}{0+1} \right) + C\\ \\ \int a(t) dt &= \frac{\cancel{2}t^2}{\cancel{t}} + 3t + C\\ \\ \int a(t) dt &= t^2 + 3t + C \end{aligned}

We know that $a(t) = v'(t)$, so

\begin{aligned} v'(0) &= (0)^2 + 3(0) + C = -4\\ \\ C &= -4 \end{aligned}

Then the velocity at time $t$ is

$v(t) = t^2 + 3t - 4$

b.) Find the distance traveled during the time period $0 \leq t \leq 3$ we know that $v(t) = t^2 + 3t - 4 = (t+4)(t-1)$, then $(t+4)(t-1) = 0$

$t = - 4$ and $t = 1$

Only $t = 1$ is in the interval $[0,3]$, thus the distance traveled is...

\begin{aligned} \int^3_0 |v(t)| dt &= \int^1_0 - v(t) dt + \int^3_1 v(t) dt\\ \\ \int^3_0 |v(t)| dt &= \int^1_0 \left( -t^2 - 3t + 4 \right) dt + \int^3_1 \left( t^2 + 3t - 4 \right) dt\\ \\ \int^3_0 |v(t)| dt &= \left[ -\frac{t^3}{3} - \frac{3t^2}{2} + 4t \right]^1_0 + \left[ \frac{t^3}{3} + \frac{3t^2}{2} - 4t \right]^3_1\\ \\ \int^3_0 |v(t)| dt &= \frac{-(1)^3}{3} - \frac{3(1)^2}{2} + 4(1) - \left[ \frac{-(0)^3}{3} - \frac{3(0)^2}{2} + 4(0) \right] + \frac{(3)^3}{3} + \frac{3(3)^2}{2} - 4(3) - \left[ \frac{(1)^3}{3} + \frac{3(1)^2}{2} - 4(1) \right]\\ \\ \int^3_0 |v(t)| dt &= \frac{-1}{3} - \frac{3}{2} + 4 + 9 + \frac{27}{2} - 12 - \frac{1}{3} - \frac{3}{2} + 4\\ \\ \int^3_0 |v(t)| dt &= \frac{89}{6} \text{ meters} \qquad \text{or} \qquad \int^3_0 |v(t)| dt = 14.83 \text{ meters} \end{aligned}