To find the impulse (I) on the bedy between $(\mathrm{t}=3)$ sec and $(\mathrm{t}=5)$ sece we need to calculate the integral of the force function ( $F(t)=3 t^2-2+2 t+$ 5) with respect to time over the glven interval.
Impulse (1) is given by:
$I=\int_8^5 F(t) d t$
Substitute ( $\mathbf{F}(t))$ :
$I=\int_3^5\left(3 t^2+2 t+5\right) d t$
We integrate each term separately:
$I=\int_3^5 3 t^2 d t+\int_3^5 2 t d t+\int_3^5 5 d t$
First, integrate $\left(3 t^2 2\right)$
$\int 3 t^2 d t=3 \cdot \frac{t_3}{3}=t^3$
Second, integrate ( $2 t$ ).
$\int 2 t d t=2 \cdot f_2^t=t^2$
Third, integrate ( 5 ):
$\int 5 d t=5 t$
Now, evaluate each integral from (3) to $(5)=$
For $\left(t^* 3\right)$
$\left[t^3\right]_3^5-5^3-3^3=125-27=98$
For $\left(t^* 2\right)=$
$\left[t^2\right]_3^5=5^2-3^2=25-9=16$
For $(5 t)$ :
$\begin{aligned} & (5 t]^5-5 \cdot 5-5 \cdot 3-25-15= \\ & 10\end{aligned}$
Sum these results:
$I=98+16+10=124$
So, the impulse on the body between $(t=3)$ sec and $(t=5)$ sec is ( 124 (kg-m/sec)).
Therefore, the correct answer is:
$124 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}$