Question

The position of a particle is given by r =3t i +2t²J +5k , where t is in second and the coefficients have the proper units for r to be in metre. Find the magnitude and direction of the velocity of the particle at t=3s.

Answer 1

Dear student
$$\begin{gathered} \overrightarrow{r}=3t\hat{i}+2t^2\hat{j}+5\hat{k} \\ Velocityofparticleis \\ \stackrel{\rightharpoonup }{v}=\frac{d\overrightarrow{r}}{dt}=\frac{d(3t\hat{i}+2t^2\hat{j}+5\hat{k})}{dt} \\ \stackrel{\rightharpoonup }{v}=(3\hat{i}+4t\hat{j}) \\ puttingt=3velocityofparticleat3secis \\ \stackrel{\rightharpoonup }{v}=(3\hat{i}+12\hat{j}) \\ magnitudeofvelocityis \\ V=\sqrt{{V_X}^2+{V_Y}^2}=\sqrt{3^2+{12}^2}=\sqrt{153}m/s \\ Xcomponentofvelocityofparticleis{V}_X=3m/s \\ Ycomponentofvelocityofparticleis{V}_Y=12m/s \\ LetanglemadebyvelocityofparticlewithXaxisis\theta so \\ \tan \theta =\frac{V_Y}{V_X}=\frac{12}{3}=4 \\ \theta ={\tan }^{-1}\left(4\right) \\ Regards \end{gathered}$$