Search any question & find its solution
Question:
Answered & Verified by Expert
A body of mass $2 \mathrm{~kg}$ travels according to the law $x(t)=p t+q t^2+r t^3$ where, $q=4 \mathrm{~ms}^{-2}, p=3 \mathrm{~ms}^{-1}$ and $r=5 \mathrm{~ms}^{-3}$. The force acting on the body at $t=2 \mathrm{~s}$ is
Options:
Solution:
1975 Upvotes
Verified Answer
The correct answer is:
$136 \mathrm{~N}$
$136 \mathrm{~N}$
As given that, mass $=2 \mathrm{~kg}$
$$
p=3 \mathrm{~m} / \mathrm{s}, q=4 \mathrm{~m} / \mathrm{s}^2, r=5 \mathrm{~m} / \mathrm{s}^3
$$
As given equation is
$$
\begin{aligned}
&x(t)=p t+q t^2+r t^3 \\
&v=\frac{d s(t)}{d t}=p+2 q t+3 r t^2 \\
&\begin{aligned}
a=\frac{d v}{d t}=\frac{d^2 x(t)}{d t^2} &=0+2 q+6 r t \\
{\left[\frac{d^2 x(t)}{d t^2}\right]_{(t=2)} } &=2 q+12 r \\
&=2 q+12 r \\
&=2 \times 4+12 \times 5 \\
&=8+60=68 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\end{aligned}
$$
Force acting on body $(\vec{F})=m a$
$$
=2 \times 68=136 \mathrm{~N} \text {. }
$$
$$
p=3 \mathrm{~m} / \mathrm{s}, q=4 \mathrm{~m} / \mathrm{s}^2, r=5 \mathrm{~m} / \mathrm{s}^3
$$
As given equation is
$$
\begin{aligned}
&x(t)=p t+q t^2+r t^3 \\
&v=\frac{d s(t)}{d t}=p+2 q t+3 r t^2 \\
&\begin{aligned}
a=\frac{d v}{d t}=\frac{d^2 x(t)}{d t^2} &=0+2 q+6 r t \\
{\left[\frac{d^2 x(t)}{d t^2}\right]_{(t=2)} } &=2 q+12 r \\
&=2 q+12 r \\
&=2 \times 4+12 \times 5 \\
&=8+60=68 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\end{aligned}
$$
Force acting on body $(\vec{F})=m a$
$$
=2 \times 68=136 \mathrm{~N} \text {. }
$$
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.