Question

Q. A car accelerates from rest at a constant rate \(\alpha\) for some time after which it decelerates at a constant rate \(\beta\) and comes to rest. If total time elapsed is \(t\), then maximum velocity acquired by car will be
A \(\left(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\right) t\)
B \(\left(\frac{\alpha^{2}-\beta^{2}}{\alpha \beta}\right) t\)
C \(\left(\frac{\alpha+\beta}{\alpha \beta}\right) t\)
D \(\left(\frac{\alpha \beta}{\alpha+\beta}\right) t\)

Answer 1

Solution:
Let maximum velocity \(=v\) Now, \(v=0+\alpha t_{1}\)
Similarly, \(0=v-\beta t_{2}\)
From the above equations we get,
$$
\begin{aligned}
&t_{1}=\frac{v}{\alpha} \quad \& \quad t_{2}=\frac{v}{\beta} \\
&t_{1}+t_{2}=t=\frac{v}{\alpha}+\frac{v}{\beta} \\
&\Rightarrow v=\frac{\alpha \beta}{\alpha+\beta} t
\end{aligned}
$$

Answer 2

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Answer 3

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