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3-2.Motion in Plane
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A particle starts from the origin at $\mathrm{t}=0$ with an initial velocity of $3.0 \hat{\mathrm{i}} \;\mathrm{m} / \mathrm{s}$ and moves in the $x-y$ plane with a constant acceleration $(6.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}) \;\mathrm{m} / \mathrm{s}^{2} .$ The $\mathrm{x}$ -coordinate of the particle at the instant when its $y-$coordinate is $32\;\mathrm{m}$ is $D$ meters. The value of $D$ is
A$50$
B$32$
C$60$
D$40$
(JEE MAIN-2020)
Solution
$\mathrm{x}=\mathrm{u}_{\mathrm{x}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2}$
$\mathrm{y}=\mathrm{u}_{\mathrm{y}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2}$
$32=0 \times t+\frac{1}{2}(4)(t)^{2}$
$t^{2}=16$
$t=4 \mathrm{sec}$
$x=3 \times 4+\frac{1}{2} \times 6 \times 4^{2}$
$=12+48=60 \mathrm{m}$
$\mathrm{y}=\mathrm{u}_{\mathrm{y}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2}$
$32=0 \times t+\frac{1}{2}(4)(t)^{2}$
$t^{2}=16$
$t=4 \mathrm{sec}$
$x=3 \times 4+\frac{1}{2} \times 6 \times 4^{2}$
$=12+48=60 \mathrm{m}$
Standard 11
Physics
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