Search any question & find its solution
Question:
Answered & Verified by Expert
A particle moves along a straight line according to the law $s=16-2 t+3 t^{3}$, where $s$ metres is the distance of the particle from a fixed point at the end of $t$ second. The acceleration of the particle at the end of $2 \mathrm{~s}$ is
Options:
Solution:
2100 Upvotes
Verified Answer
The correct answer is:
$36 \mathrm{~m} / \mathrm{s}^{2}$
Given, $s=16-2 t+3 t^{3}$
$$
\begin{array}{lc}
\Rightarrow & \frac{d s}{d t}=-2+9 t^{2} \\
\Rightarrow \quad \frac{d^{2} s}{d t^{2}}=18 t
\end{array}
$$
Now, the acceleration of the particle at the end of $t=2 \mathrm{~s}$ is
$f=\frac{d^{2} s}{d t^{2}}=18 \times 2$
$=36 \mathrm{~m} / \mathrm{s}^{2}$
$$
\begin{array}{lc}
\Rightarrow & \frac{d s}{d t}=-2+9 t^{2} \\
\Rightarrow \quad \frac{d^{2} s}{d t^{2}}=18 t
\end{array}
$$
Now, the acceleration of the particle at the end of $t=2 \mathrm{~s}$ is
$f=\frac{d^{2} s}{d t^{2}}=18 \times 2$
$=36 \mathrm{~m} / \mathrm{s}^{2}$
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.