Solve the problem.
-A certain object moves in such a way that its velocity (in $\mathrm{m} / \mathrm{s}$ ) after time $\mathrm{t}$ (in $\mathrm{s}$ ) is given by $v=t^{2}+3 t+10$. Find the distance traveled during the first four seconds.
A) $38.0 \mathrm{~m}$
B) $45.3 \mathrm{~m}$
C) $85.3 \mathrm{~m}$
D) $64.0 \mathrm{~m}$

Question

Quizplus · Accepted Answer

The distance traveled is the integral of velocity over time. Integrate $v = t^2 + 3t + 10$ from 0 to 4 to get $\int_0^4 (t^2 + 3t + 10) \, dt = \left[\frac{t^3}{3} + \frac{3t^2}{2} + 10t\right]_0^4 = 85.3 \, \mathrm{m}$.

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