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 $(6.0 \hat{i}+4.0 \hat{j}) \mathrm{m} / \mathrm{s}^{2}$. The $x$-coordinate of the particle at the instant when its $y$-coordinate is 32 m is D meters. The value of D is

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Kinematics

A particle starts from the origin at

t = 0

with an initial velocity of

3.0 \hat{i} m / s

and moves in the

x

y

plane with a constant acceleration

(6.0 \hat{i} + 4.0 \hat{j}) m / 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

(A)40
(B)60✓ Correct
(C)32
(D)50

Answer: (B)

Solution

Treat the two directions separately. Vertically the particle starts with no

y

-velocity and accelerates at

4 m/s^{2}

, so

32 = \frac{1}{2} (4) t^{2}

gives

t = 4 s

. Horizontally it starts at

3 m/s

and accelerates at

6 m/s^{2}

, so

x = 3 (4) + \frac{1}{2} (6) (4)^{2} = 12 + 48 = 60 m

. The

x

-coordinate when

y = 32 m

60

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option

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