A particle starts from origin at $\mathrm{t}=0$ with a velocity $5 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}$ and moves in $\mathrm{x}-\mathrm{y}$ plane under action of a force which produces a constant acceleration of $(3 \hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}^{2}$. If the $x$-coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is $\sqrt{\alpha} \mathrm{m} / \mathrm{s}$. The value of $\alpha$ is ______.

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Kinematics

A particle starts from origin at

t = 0

with a velocity

5 \hat{i} m / s

and moves in

x - y

plane under action of a force which produces a constant acceleration of

(3 \hat{i} + 2 \hat{j}) m / s^{2}

. If the

x

-coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is

α m / s

. The value of

α

is ______.

Answer: 673

Solution

Taking

x

u_{x} = 5

m/s,

a_{x} = 3

m/s²,

x = 84

v_{x}^{2} = u_{x}^{2} + 2 a_{x} x = 25 + 504 = 529 \Rightarrow v_{x} = 23 m/s

Time to reach this point:

v_{x} = u_{x} + a_{x} t \Rightarrow 23 = 5 + 3 t \Rightarrow t = 6

s.

In

y

u_{y} = 0, a_{y} = 2

m/s², so

v_{y} = a_{y} t = 12

m/s.

v^{2} = v_{x}^{2} + v_{y}^{2} = 2 3^{2} + 1 2^{2} = 529 + 144 = 673

α = 673

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Integer answer

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