Search any question & find its solution
Question:
Answered & Verified by Expert
A body is moving with velocity $30 \mathrm{~m} / \mathrm{s}$ towards east. After $10 \mathrm{~s}$ its velocity becomes $40 \mathrm{~m} / \mathrm{s}$ towards north. The average acceleration of the body is
Options:
Solution:
1538 Upvotes
Verified Answer
The correct answer is:
$5 \mathrm{~m} / \mathrm{s}^2$
$$
\begin{aligned}
& \text { Average acceleration }=\frac{\text { Change in velocity }}{\text { Total time }} \\
& \qquad \begin{aligned}
a & =\frac{\left|\overrightarrow{\mathbf{v}}_f-\overrightarrow{\mathbf{v}}_i\right|}{\Delta t} \\
& =\frac{\sqrt{30^2+40^2}}{10}=\frac{\sqrt{900+1600}}{10} \\
& =5 \mathrm{~ms}^{-2}
\end{aligned}
\end{aligned}
$$
\begin{aligned}
& \text { Average acceleration }=\frac{\text { Change in velocity }}{\text { Total time }} \\
& \qquad \begin{aligned}
a & =\frac{\left|\overrightarrow{\mathbf{v}}_f-\overrightarrow{\mathbf{v}}_i\right|}{\Delta t} \\
& =\frac{\sqrt{30^2+40^2}}{10}=\frac{\sqrt{900+1600}}{10} \\
& =5 \mathrm{~ms}^{-2}
\end{aligned}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.