Question

A point initially at rest moves along x axis. Its deceleration varies with time as a= 5t + 6. Calculate distance travelled in 2 second

Answer 1

Dear student

$$\begin{gathered} a=-(5t+6) \\ \frac{dv}{dt}=-5t-6 \\ {\int }_0^{v}dv={\int }_0^t(-5t-6)dt \\ v=\frac{-5t^2}{2}-6t \\ forv=0 \\ t=0thisconfirmsthatbodyismovinginstraightpathsodis\tan ce=displacement \\ \frac{dx}{dt}=(\frac{-5t^2}{2}-6t) \\ {\int }_0^{x}dx={\int }_0^2(\frac{-5t^2}{2}-6t)dt \\ x=\frac{-5(t^3{{)}_0}^2}{6}-6\frac{(t^2{{)}_0}^2}{2}=\frac{-5\times 8}{6}-3\left(4\right)=\frac{-20}{3}-12=\frac{-20-36}{3}=\frac{-56}{3}m= \\ sodis\tan ceis56/3m \\ Regards \end{gathered}$$