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JEE MAIN 2020
09-01-2020 S2
Question
A particle starts from the origin at $t=0$ with an initial velocity of $3.0 \hat{i} \mathrm{~m} / \mathrm{s}$ 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 $x$-coordinate of the particle at the instant when its $y$-coordinate is 32 m is D meters. The value of D is
Select the correct option:
A
60
B
32
C
40
D
50
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \frac{1}{2} \times 4 \times t^2=32 \\ & t=4 s \\ & x=3 \times 4+\frac{1}{2} \times 6 \times 4^2=60\end{aligned}$
Question Tags
JEE Main
Physics
Easy
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