Solve the problem.
-A certain object moves in such a way that its velocity (in $\mathrm { m } / \mathrm { s }$ ) after time $t$ (in s) is given by $\mathrm { v } = \mathrm { t } ^ { 2 } + 2 \mathrm { t } + 12$. Find the distance traveled during the first four seconds by evaluating $\int _ { 0 } ^ { 4 } \left( t ^ { 2 } + 2 t + 12 \right)$ dt. Round to the nearest tenth.
A) $64.0 \mathrm {~m}$
B) $85.3 \mathrm {~m}$
C) $37.3 \mathrm {~m}$
D) $36.0 \mathrm {~m}$

Question

Quizplus · Accepted Answer

To find the distance traveled, we integrate the velocity function over the given time interval. The integral of $v = t^2 + 2t + 12$ from 0 to 4 is $\frac{1}{3}t^3 + t^2 + 12t$ evaluated from 0 to 4. Plugging in the limits, we get $\frac{1}{3}(4)^3 + (4)^2 + 12(4) - (\frac{1}{3}(0)^3 + (0)^2 + 12(0)) = \frac{64}{3} + 16 + 48 = \frac{64 + 48*3 + 16*3}{3} = \frac{64 + 144 + 48}{3} = \frac{256}{3} = 85.3$. Therefore, the distance traveled is 85.3 meters.

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