Answer
a. i) $33.2$, ii) $33.2$, iii) $33.2$ units.
b. Yes.
Work Step by Step
a. Given $\frac{ds}{dt}=v=9.8t-3$, we have $s(t)=\int(9.8t-3)dt=4.9t^2-3t+C$ where $C$ is a constant.
i) As $s(0)=5$, we have $4.9(0)^2-3(0)+C=5$ and $C=5$. The displacement is $\Delta s=s(3)-s(1)=4.9(3)^2-3(3)-(4.9(1)^2-3(1))=33.2$
ii) As $s(0)=-2$, we have $4.9(0)^2-3(0)+C=-2$ and $C=-2$. The displacement is $\Delta s=s(3)-s(1)=4.9(3)^2-3(3)-(4.9(1)^2-3(1))=33.2$
iii) As $s(0)=s_0$, we have $4.9(0)^2-3(0)+C=s_0$ and $C=s_0$. The displacement is $\Delta s=s(3)-s(1)=4.9(3)^2-3(3)-(4.9(1)^2-3(1))=33.2$ units.
b. The answer to this question is an extension of the above results. The displacement $\Delta s$ will not depend on the initial condition $s_0$ as the constant will be cancelled out. So the answer is Yes.
