JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Kinematics

The co-ordinates of a particle moving in

x - y

plane are given by :

x = 2 + 4 t, y = 3 t + 8 t^{2}

.
The motion of the particle is :

(A)non-uniformly accelerated.
(B)uniformly accelerated having motion along a straight line.
(C)uniform motion along a straight line.
(D)uniformly accelerated having motion along a parabolic path.✓ Correct

Answer: (D)

Solution

From

x = 2 + 4 t

v_{x} = 4

a_{x} = 0

.
From

y = 3 t + 8 t^{2}

v_{y} = 3 + 16 t

a_{y} = 16

.
The acceleration

(0, 16)

is constant and non-zero

\Rightarrow

uniformly accelerated motion.

Eliminating

t = \frac{x - 2}{4}

y = 3 (\frac{x - 2}{4}) + 8 (\frac{x - 2}{4})^{2},

a quadratic in

x

— a parabolic path. Option D.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option

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