Question

In the previous problem, if the particle occupies a position $x=7m$ at $t=1s$, then obtain an expression for the instantaneous displacement of the particle.

Answer 1

Again, we can use the idea that displacement is the integration of velocity w.r.t. time.
So, $x=\int vdt=\int (2t+5)dt=\frac{2t^2}{2}+5t+c=t^2+5t+c$
where $c$ is the constant of integration. Its value can be determined by using the given condition. (As particular details have been given about the particle.)
At $t=1s,x=7m=7=1^2+5\times 1\Rightarrow c=1m$
Hence, the expression becomes $x=t^2+5t+1$.