Answer :
Solution.
Given speed as a function of time as shown below
[tex]s(t)=3t-\frac{1}{4}t^3[/tex][tex]\begin{gathered} Recall\text{ that speed = }\frac{\text{distance}}{time\text{ taken}} \\ That\text{ is, } \\ v(t)=\frac{ds}{dt} \\ ds=v(t)dt \\ Integrating\text{ both sides} \end{gathered}[/tex][tex]\begin{gathered} \int ds=\int_a^bv(t)dt \\ s=\int_a^bv(t)dt \\ Where\text{ s = distance, v\lparen t\rparen =speed function = 3t-}\frac{1}{4}t^3 \\ Thus,\text{ distance travelled after 4 secs can be represented by } \\ s=\int_0^4(3t-\frac{1}{4}t^3)dt \\ \end{gathered}[/tex][tex]\begin{gathered} s=\int_0^43tdt-\frac{1}{4}t^3dt \\ s=\lbrack\frac{3t^2}{2}-\frac{1}{4}\frac{t^4}{4}\rbrack_{0\text{ }}^4 \end{gathered}[/tex][tex]\begin{gathered} s=(\frac{3t^2}{2}-\frac{1}{16}t^4)_0^4 \\ s=\frac{3(4^2)}{2}-\frac{1}{16}(4^4) \\ s=24-16 \\ s=8 \end{gathered}[/tex]Thus, the disthe distance traveled by the object from the start to the end of the 4th second is 8 units