Question

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 $\left. m , y _{0} m \right)$. The values of $t$ and $y _{0}$, are respectively

• A. $4 \,s$ and $52\, m$
• B. $2\, s$ and $24\, m$
• C. $2\, s$ and $18\, m$
• D. $5\, s$ and $25\, m$

Answer 1

The Correct Option is C

To determine the values of $t$ and $y_0$ for the particle moving in the $x-y$ plane, we'll break the problem down step-by-step.

1. Given that the particle starts at the origin $(0,0)$ with initial velocity $5 \hat{j} \, \text{ms}^{-1}$ and is subjected to an acceleration of $(10 \hat{i} + 4 \hat{j}) \, \text{ms}^{-2}$.
2. For the $x$-coordinate:
   • Use the equation of motion: $x = u_x t + \frac{1}{2} a_x t^2$.
   • Since the initial velocity in the $x$-direction is $0$, the equation simplifies to: $x = \frac{1}{2} a_x t^2$.
   • Substituting the values, we have $20 = \frac{1}{2} \times 10 \times t^2$.
   • Simplifying gives: $20 = 5t^2$, hence $t^2 = 4$ leading to $t = 2 \, \text{s}$.
3. For the $y$-coordinate:
   • Use the equation of motion: $y = u_y t + \frac{1}{2} a_y t^2$.
   • The initial velocity in the $y$-direction is $5 \, \text{ms}^{-1}$, giving: $y = 5t + \frac{1}{2} \times 4 \times t^2$.
   • Substituting $t = 2 \, \text{s}$, we have $y = 5 \times 2 + \frac{1}{2} \times 4 \times 2^2$.
   • Calculating gives: $y = 10 + 8 = 18 \, \text{m}$.

Thus, the values of $t$ and $y_0$ are $2 \, \text{s}$ and $18 \, \text{m}$, respectively, which corresponds to the correct option.