Starting from the origin at time $t=0$, with initial velocity $5 \hat{j} \mathrm{~ms}^{-1}$, a particle moves in the $x$ - y plane with a constant acceleration of $(10 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{ms}^{-2}$. At time t , its coordinates are ( 20 $\mathrm{m}, \mathrm{y}_{0} \mathrm{~m}$ ). The values of t and $\mathrm{y}_{0}$ are, respectively:

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Kinematics

Starting from the origin at time

t = 0

, with initial velocity

5 \hat{j} ms^{- 1}

, a particle moves in the

x

- y plane with a constant acceleration of

(10 \hat{i} + 4 \hat{j}) ms^{- 2}

. At time t , its coordinates are ( 20

m, y_{0} m

). The values of t and

y_{0}

are, respectively:

Answer: (D)

With constant acceleration the displacement is

s = u t + \frac{1}{2} a t^{2}

. Here

u = 5 \hat{j}

and

a = 10 \hat{i} + 4 \hat{j}

, so

s = 5 t \hat{j} + \frac{1}{2} (10 \hat{i} + 4 \hat{j}) t^{2} = 5 t^{2} \hat{i} + (5 t + 2 t^{2}) \hat{j}

. Matching the

x

-coordinate to

20

m gives

5 t^{2} = 20

, so

t = 2

s. Putting this into the

y

-coordinate,

y_{0} = 5 (2) + 2 (2)^{2} = 10 + 8 = 18

m. The particle is at that point after

2

s with

y_{0} = 18

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option