Physics - Kinematics Question with Solution | TestHub

The velocity vector is found by differentiating the position vector with respect to time: $\vec{v}(t)=\frac{d\vec{r}}{dt}=(3t^2-3)\hat{i}+(8t)\hat{j}$. The acceleration vector is found by differentiating the velocity vector with respect to time: $\vec{a}(t)=\frac{d\vec{v}}{dt}=(6t)\hat{i}+(8)\hat{j}$. At $t=1\text{s}$, the acceleration vector is $\vec{a}(1)=(6(1))\hat{i}+(8)\hat{j}=6\hat{i}+8\hat{j}$. The magnitude of the acceleration is $|\vec{a}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10{\text{m/s}}^2$.