Question 4.1: A particle moves over a path such that the components of its......
A particle moves over a path such that the components of its position with respect to an origin of coordinates are given as a function of time by:
\begin{array}{c}{{x=-t^{2}+12t+5}}\\ {{y=-2\,t^{2}+16t+10}}\end{array}where t is in seconds and x and y are in meters. (a) Find the particle’s position vector \vec{r} as a function of time, and find its magnitude and direction at t = 6 s. (b) Find the particle’s velocity vector \vec{v} as a function of time, and find its magnitude and direction at t = 6 s. (c) Find the particle’s acceleration vector \vec{a} as a function of time, and find its magnitude and direction at t = 6 s.
(a) The position vector is given at time t by:
{\vec{r}}=x{\vec{{i}}}+y{\vec{{j}}}=(-t^{2}+12t+5){\vec{{i}}}+(-2t^{2}+16t+10){\vec{{j}}}Figure 4.6 shows the variation of x and y as a function of time. At t = 6 s we have:
{\vec{r}}\,=x \ \vec{i}+y{\vec{j}}\,=41{\vec{\ i}}\,+34{\vec{j}}The magnitude of \vec{r} is:
r={\sqrt{x^{2}+y^{2}}}={\sqrt{41^{2}+34^{2}}}=53.26\,{\mathrm{m}}The angle θ between \vec{r} and the direction of increasing x is:
\theta=\tan^{-1}\left({\frac{y}{x}}\right)=\tan^{-1}\left({\frac{34 \ {\mathrm{m}}}{41\ {\mathrm{m}}}}\right)=\tan^{-1}(0.83)=39.7^{\circ}Figure 4.7 shows the path of the particle in the xy plane, and also shows its position vector \vec{r} at t = 6 s.
(b) The velocity components along the x and y axes are:
\ v_{x}={\frac{d x}{d t}}={\frac{d}{d t}}(-t^{2}+12\,t+5)=-2\,t+12,
v_{y}={\frac{d y}{d t}}={\frac{d}{d t}}(-2\,t^{2}+16\,t+10)=-4\,t+16
At t = 6 s the components of \vec{v} are:
v_{x}=0,\quad v_{y}=-8\,\mathrm{m/s}That is, \vec{v} = −8 \vec{j} . The magnitude of \vec{v} at this time is:
v={\sqrt{v_{x}^{2}+v_{y}^{2}}}={\sqrt{0+(-8\,{\mathrm{m/s}})^{2}}}=8\,{\mathrm{m/s}}Hence, the angle θ between \vec{v} and the direction of increasing x is:
\theta=\tan^{-1}\left({\frac{v_{y}}{v_{x}}}\right)=\tan^{-1}\left({\frac{-8\;{\mathrm{m/s}}}{0\;{\mathrm{m/s}}}}\right)=\tan^{-1}(-\infty)=270^{\circ}where we used the fact that \vec{v} = −8 \vec{j} is a downward vector and its angle should be measured in a counterclockwise sense from the direction of increasing x, see the Fig. 4.7.
(c) The components of the acceleration along the x and y axes are:
a_{x}={\frac{d v_{x}}{d t}}={\frac{d}{d t}}(-2t+12)=-2\,{\mathrm{m/s}}^{2},
a_{y}={\frac{d v_{y}}{d t}}={\frac{d}{d t}}(-4t+16)=-4\,{\mathrm{m/s}}^{2}
We see that the acceleration does not vary with time, i.e. it is a constant. We define the magnitude and direction of \vec{a} as follows:
a={\sqrt{a_{x}^{2}+a_{y}^{2}}}={\sqrt{(-2\,{m/s^{2}})^{2}+(-4\,\operatorname{m/s}^{2})^{2}}}={\sqrt{20\,(\operatorname{m/s}^{2})^{2}}}=4.47\,\mathrm{m/s}^{2}
\theta=\tan^{-1}\left({\frac{a_{y}}{a_{x}}}\right)=\tan^{-1}\left({\frac{-4{\mathrm{~m/s}}^{2}}{-2{\mathrm{~m/s}}^{2}}}\right)=180^{\circ}+\tan^{-1}(2)
= 180° + 63.4° = 243.4°
where we used Table 2.1 to calculate θ for a negative a_{x} and a_{y} .
Table 2.1 Calculating θ from Φ according to the signs of A_{x} and A_{y}
| Sign of A_{x} | Sign of A_{y} | Quadrant | Angle θ |
| + | + | I | θ = Φ |
| – | + | II | θ = 180° − Φ |
| – | – | III | θ = 180° + Φ |
| + | – | IV | θ = 360° − Φ |
