Question 10
The position vector (m) of a stone $t$ seconds after it is thrown is
$$\begin{aligned}\boldsymbol{r}(t)=\ &18\cos(25^{\circ})\,t\,\hat{\boldsymbol{\imath}}\\&+\left(18\sin(25^{\circ})\,t-4.9t^{2}\right)\hat{\boldsymbol{\jmath}}\end{aligned}$$
Determine the speed (m s$^{-1}$) of the stone when $t=2$.
Not quite. You gave the magnitude of the acceleration, $\left\lvert\boldsymbol{a}\right\rvert=9.8$.
Not quite. You assumed the speed stays at the launch speed, $\left\lvert\boldsymbol{v}(0)\right\rvert=18$.
That’s it. Step through the hints below to see why.
Not quite. Your calculator was in radian mode, so it evaluated $\cos(25)$ and $\sin(25)$ with 25 in radians.
Differentiate $\boldsymbol{r}$.
$$\begin{aligned}\boldsymbol{v}(t)=\ &18\cos(25^{\circ})\,\hat{\boldsymbol{\imath}}\\&+\left(18\sin(25^{\circ})-9.8t\right)\hat{\boldsymbol{\jmath}}\end{aligned}$$
Find $\left\lvert\boldsymbol{v}(2)\right\rvert$ in degree mode.
In degree mode, enter √((18cos(25))^2+(18sin(25)−9.8×2)^2). Output: $20.2475$.
$$\begin{aligned}\boldsymbol{v}(t)=\ &18\cos(25^{\circ})\,\hat{\boldsymbol{\imath}}\\&+\left(18\sin(25^{\circ})-9.8t\right)\hat{\boldsymbol{\jmath}}\end{aligned}$$
The speed is $20.25$ m s$^{-1}$, option (C).
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it