I'm having a difficult time figuring out this problem. I'm thinking that you will need to solve the anti-derivative to get the velocity, and then solve for C, using the initial velocity. Next, wouldn't you solve for the integral of velocity, using the interval given? I'm getting a really long, complicated velocity function so I'm not sure if I'm doing it right?
$\endgroup$ 3If a particle is traveling with an acceleration of $a(t) = \frac{t}{\sqrt{t+2}}$, for $1 \leq t \leq 5$, when its initial velocity is $2$, determine the total distance traveled during the time interval.
1 Answer
$\begingroup$The integral of the acceleration function is the velocity function:
$$v(t) = \int \frac{t}{\sqrt{t+2}} \, dt = \frac{2}{3}(t - 4) \sqrt{t + 2} + C$$
To solve for $C$, we use initial time $t=1$ and initial velocity $2$ and solve $\dfrac{2}{3}(1 - 4) \sqrt{1 + 2} + C = 2$ which comes out to $C = 2 (1 + \sqrt{3})$.
Finally, to get the distance traveled, we can take the integral of the velocity function from $t=1$ to $5$:
$$p(5) - p(1) = \int_{1}^{5} \frac{2}{3}(t - 4) \sqrt{t + 2} + 2 (1 + \sqrt{3}) \, dt = \dfrac{4}{5} (10 + 17 \sqrt{3} - 7 \sqrt{7}) = 16.74$$
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