TS EAMCET · Physics · Motion In One Dimension
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
- A \(\frac{27 \alpha+9 \beta}{26 \alpha+6 \beta}\)
- B \(\frac{9 \alpha+3 \beta}{18 \alpha+4 \beta}\)
- C \(\frac{13 \alpha+4 \beta}{27 \alpha+6 \beta}\)
- D \(\frac{26 \alpha+8 \beta}{9 \alpha+3 \beta}\)
Answer & Solution
Correct Answer
(C) \(\frac{13 \alpha+4 \beta}{27 \alpha+6 \beta}\)
Step-by-step Solution
Detailed explanation
Average Velocity \(V_1=\frac{\text { Displacement }}{\text { time }}\) We have \(x=\alpha t^3+\beta t^2+\gamma\)…
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