Position of a particle is given by r=3.0 t hat{i}+2.0 t^{2} hat{j}+5.0 hat{k} where t is in seconds

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3-2.Motion in Plane

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The position of a particle is given by

$r=3.0 t \hat{i}+2.0 t^{2} \hat{j}+5.0 \hat{k}$

where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres.

$(a)$ Find $v (t)$ and $a (t)$ of the particle.

$(b)$ Find the magnitude and direction of $v (t)$ at $t=1.0 s$

Option A

Option B

Option C

Option D

Solution

$v (t)=\frac{ d r }{ d t}=\frac{ d }{ d t}\left(3.0 t \hat{ i }+2.0 t^{2} \hat{ j }+5.0 \hat{ k }\right)$

$=3.0 \hat{ i }+4.0 t \hat{ j }$

$a (t)=\frac{ d v }{ d t}=+4.0 \hat{ j }$

$ a=4.0 m s ^{-2}$ along $y-$ direction

At $t=1.0 s , \quad v =3.0 \hat{ i }+4.0 \hat{ j }$

It's magnitude is $v=\sqrt{3^{2}+4^{2}}=5.0 m s ^{-1}$

and direction is $\theta=\tan ^{-1}\left(\frac{v_{y}}{v_{x}}\right)=\tan ^{-1}\left(\frac{4}{3}\right) \cong 53^{\circ}$ with $x$ -axis.

Standard 11

Physics

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