Motion In A Plane Ques 22
22. A body of $3 $ $kg$ moves in the $X Y$ plane under the action of a force given by $6 t^{2} \hat{i}+4 t \hat{j}$.
Assuming that the body is at rest at time $t=0$, the velocity of the body at $t=3 s$ is
[2002]
(a) $6 \hat{i}+6 \hat{j}$
(b) $18 \hat{i}+6 \hat{j}$
(c) $18 \hat{i}+12 \hat{j}$
(d) $12 \hat{i}+18 \hat{j}$
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Answer:
Correct Answer: 22.(b)
Solution:
- (b) $\vec{F}=6 t \hat{i}+4 t \hat{j}$
$F_x=6 t, F_y=4 t$
$a_x=\frac{6 t}{3}=2 t, a_y=\frac{4 t}{3}$
$\frac{d v_x}{d t}=2 t^{2}$
$\int_0^{v_x} d V_x=\int_0^{t} 2 t^{2} d t$
$\Rightarrow v_x=\frac{2}{3} t^{3}=\frac{2}{3} \cdot 3^{3}=18$
$ \frac{d V_y}{d t}=\frac{4}{3} t \Rightarrow \int_0^{V_y} d V_y=\frac{4}{3} \int_0^{t} t d t $
$\Rightarrow \quad V_y=\frac{4}{3} \frac{t^{2}}{2}=\frac{4}{3} \cdot \frac{9}{2}=6$
$\Rightarrow V=18 \vec{t}+6 \vec{T}$
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