Question

Starting from the origin at time $\mathrm{t}=0$, with an initial velocity $5\widehat{\mathrm{j}}\mathrm{m}/\mathrm{s}$, a particle moves in the $\mathrm{xy}$ plane with a constant acceleration of $(10\widehat{\mathrm{i}}+4\widehat{\mathrm{j}})\mathrm{m}/{\mathrm{s}}^2$.At time t, its coordinates are $\left(20,y_{\circ }\right)$.The values of $t$ and $y_{\circ }$ are, respectively:

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

Answer 1

The correct option is B

$2\mathrm{s}\mathrm{and}18\mathrm{m}$

Step 1: Given data

The initial velocity $\overrightarrow{u}=5\hat{j}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Acceleration of particle $\overrightarrow{a=}(10\widehat{i}+4\widehat{j})m/s^2$

Position coordinate at time $\text{'}t\text{'}$= $\left(20,y_{\circ }\right)$

Step 2: Formula used

The second equation of motion is given by-

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

Step 3: Calculating the values of $t$ and $y_{\circ }$

From the formula,

$s=ut+\frac{1}{2}a{t}^2$.

On putting the values and further solving,

$$\begin{gathered} 20\widehat{i}+y_{\circ }\widehat{j}=5\widehat{j}t+\frac{1}{2}(10\widehat{i}+4\widehat{j})t^2 \\ 20\widehat{i}+y_{\circ }\widehat{j}=5\widehat{j}t+5t^2\hat{i}+2t^2\hat{j} \\ 20\widehat{i}+y_{\circ }\widehat{j}=5t^2\hat{i}+\left(5t+2t^2\right)\hat{j} \end{gathered}$$

Compare $X$ component of the position,

$\begin{array}{l}20=5t^2\\ t^2=4\\ t=2s\end{array}$

Compare $Y$ component of the position,

$\begin{array}{l}{y}_{\circ }=5t+2t^2\\ y_{\circ }=5\times 2+2\times 2^2\\ y_{\circ }=18\end{array}$

Thus, the value of $t$ and $y_{\circ }$ are $2s,18m$ respectively.

Hence, option B is the correct answer.