The position of a particle is given by r = 3.0t i -2.0t2 j + 4.0 k m. where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the v and a of the particle?
(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?
\( \vec{r}(t) = 3.0 t \,\hat{i} - 2.0 t^{2} \,\hat{j} + 4.0 \,\hat{k} \;\text{m} \)
Velocity is the time derivative of position:
\( \vec{v}(t) = \dfrac{d\vec{r}}{dt} \) \( = \dfrac{d}{dt} (3.0 t \,\hat{i} - 2.0 t^{2} \,\hat{j} + 4.0 \,\hat{k}) \) \( = 3.0 \,\hat{i} - 4.0 t \,\hat{j} + 0 \,\hat{k} \) \( \Rightarrow \vec{v}(t) = (3.0\,\hat{i} - 4.0 t\,\hat{j}) \,\text{m s}^{-1} \)
Acceleration is the time derivative of velocity:
\( \vec{a}(t) = \dfrac{d\vec{v}}{dt} \) \( = \dfrac{d}{dt} (3.0 \,\hat{i} - 4.0 t \,\hat{j}) \) \( = 0 \,\hat{i} - 4.0 \,\hat{j} \) \( \Rightarrow \vec{a}(t) = -4.0 \,\hat{j} \,\text{m s}^{-2} \)
(a) \( \vec{v}(t) = (3.0\,\hat{i} - 4.0 t\,\hat{j}) \,\text{m s}^{-1} \), \( \vec{a}(t) = -4.0 \,\hat{j} \,\text{m s}^{-2} \).
\( \vec{v}(2) = 3.0\,\hat{i} - 4.0 (2)\,\hat{j} = 3.0\,\hat{i} - 8.0\,\hat{j} \,\text{m s}^{-1} \)
\( v = |\vec{v}(2)| = \sqrt{3.0^{2} + (-8.0)^{2}} = \sqrt{9 + 64} = \sqrt{73} \approx 8.54 \,\text{m s}^{-1} \)
Let \( \theta \) be the angle that \( \vec{v}(2) \) makes with the +x axis in the x–y plane. Then
\( \tan \theta = \dfrac{v_{y}}{v_{x}} = \dfrac{-8.0}{3.0} \approx -2.67 \)
Hence
\( \theta = \tan^{-1}(-2.67) \approx -69.4^{\circ} \)
This means the velocity vector makes an angle of \( 69.4^{\circ} \) below the +x axis (in the fourth quadrant).
(b) Magnitude of velocity at \( t = 2.0 \,\text{s} \): \( v \approx 8.54 \,\text{m s}^{-1} \). Direction: \( 69.4^{\circ} \) below the +x axis.