Q77 / 90 · JEE Main 2009 · Physics
A particle has an initial velocity $3 \hat{i}+4 \hat{j}$ and an acceleration of $0.4 \hat{i}+0.3 \hat{j}$. Its speed after 10 s is
- A10 units
- Correct answerB$7 \sqrt{2}$ units
- C7 units
- D$8.5$ units
Solution
$$
\begin{aligned}
\overrightarrow{\mathrm{u}} & =3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} ; \overrightarrow{\mathrm{a}}=0.4 \hat{\mathrm{i}}+0.3 \hat{\mathrm{j}} \\
\overrightarrow{\mathrm{u}} & =\overrightarrow{\mathrm{u}}+\overrightarrow{\mathrm{a}} \mathrm{t} \\
& =3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+(0.4 \hat{\mathrm{i}}+0.3 \hat{\mathrm{j}}) 10=3 \hat{i}+\hat{\mathrm{j}}+4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}=7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}
\end{aligned}
$$
Speed is $\sqrt{7^2+7^2}=7 \sqrt{2}$ units
Hence, the correct answer is option: 'B'.