Search any question & find its solution
Question:
Answered & Verified by Expert
A particle moves along a straight line such that its displacement ' $x$ ' varies with time ' $t$ ' as $x=\alpha t^3+\beta t^2+\gamma$, where $\alpha, \beta, \gamma$ are constants. $V_1$ is the average velocity of the particle during its journey between $t=1 \mathrm{~s}$ and $t=3 \mathrm{~s} . \mathrm{V}_2$ is the instantaneous velocity of the particle at $t=3 \mathrm{~s}$. The ratio $\frac{\mathrm{V}_1}{\mathrm{~V}_2}$ is
Options:
Solution:
1240 Upvotes
Verified Answer
The correct answer is:
$\frac{13 \alpha+4 \beta}{27 \alpha+6 \beta}$
Average Velocity $V_1=\frac{\text { Displacement }}{\text { time }}$
We have $x=\alpha t^3+\beta t^2+\gamma$
$$
\begin{aligned}
& \mathrm{V}_1=\frac{\left[\alpha(3)^3+\beta(3)^2+\gamma\right]-\left[\alpha(1)^3+\beta(1)^2+\gamma\right]}{3-1} \\
& \mathrm{~V}_1=\frac{26 \alpha+10 \beta}{2} \\
& \mathrm{~V}_1=13 \alpha+5 \beta
\end{aligned}
$$
Instantaneous Velocity $\mathrm{V}_2=\left.\frac{\mathrm{dx}}{\mathrm{dt}}\right|_{\mathrm{t}=3}$
$$
=\left[3 \alpha \mathrm{t}^2+2 \beta \mathrm{t}\right]_{\mathrm{t}=3}=27 \alpha+6 \beta
$$
The ratio $\frac{V_1}{V_2}=\frac{13 \alpha+5 \beta}{27 \alpha+6 \beta}$
We have $x=\alpha t^3+\beta t^2+\gamma$
$$
\begin{aligned}
& \mathrm{V}_1=\frac{\left[\alpha(3)^3+\beta(3)^2+\gamma\right]-\left[\alpha(1)^3+\beta(1)^2+\gamma\right]}{3-1} \\
& \mathrm{~V}_1=\frac{26 \alpha+10 \beta}{2} \\
& \mathrm{~V}_1=13 \alpha+5 \beta
\end{aligned}
$$
Instantaneous Velocity $\mathrm{V}_2=\left.\frac{\mathrm{dx}}{\mathrm{dt}}\right|_{\mathrm{t}=3}$
$$
=\left[3 \alpha \mathrm{t}^2+2 \beta \mathrm{t}\right]_{\mathrm{t}=3}=27 \alpha+6 \beta
$$
The ratio $\frac{V_1}{V_2}=\frac{13 \alpha+5 \beta}{27 \alpha+6 \beta}$
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.