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Motion in a Plane

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Question : 20 of 32
Marks: +1, -0
The position of a particle is given by r=3.0ti^−2.0t2j^+4.0k^mr=3.0 t \hat{i}-2.0 t^{2} \hat{j}+4.0 \hat{k} \mathrm{m} where tt is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the v⃗\vec{v} and a⃗\vec{a} of the particle?
(b) What is the magnitude and direction of velocity of the particle at t=2.0st = 2.0 \mathrm{s}?
Solution:  
(a) Velocity,
v⃗=dr⃗dt\vec{v}= \frac{d\vec{r}}{dt} =ddt(3.0ti^−2.0t2j^+4.0k^)= \frac{d}{dt}(3.0 t \hat{i}-2.0 t^{2} \hat{j}+4.0 \hat{k}) =[3.0i^−4.0tj^]ms−1=[3.0 \hat{i}-4.0 t \hat{j}] \mathrm{m} \mathrm{s}^{-1}
Acceleration, a⃗=dv⃗dt\vec{a}= \frac{d\vec{v}}{dt} =ddt(3.0i^−4.0tj^)=0−4.0j^ms−2= \frac{d}{dt}(3.0 \hat{i}-4.0 t \hat{j})=0-4.0 \hat{j} \mathrm{m} \mathrm{s}^{-2}
(b) At time t=2s,v⃗=3.0i⃗−4.0×2j⃗t=2 \mathrm{s}, \vec{v}=3.0 \vec{i}-4.0 \times 2 \vec{j} =3.0i⃗−8.0j⃗=3.0 \vec{i}-8.0 \vec{j}
∴v=(3.0)2+(−8)2=73=8.54ms−1\therefore v = \sqrt{(3.0)^{2}+(-8)^{2}}=\sqrt{73}=8.54 \mathrm{m} \mathrm{s}^{-1}
If θ\theta is the angle which makes with x-axis, then
tan⁡θ=vyvx=−83=−2.667\tan \theta= \frac{v_{y}}{v_{x}}= \frac{-8}{3}=-2.667
∴θ=tan⁡−1(−2.667)≃70∘\therefore\theta=\tan ^{-1}(-2.667) \simeq 70^{\circ} with xx -axis
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