Question

A car accelerates from rest at a constant rate a for some time after which it decelerates at a constant speed $\beta$ to come to rest. If the total time elapsed is ,, the maximum velocity acquired and total distance travelled by the car are given by
(a) $\left(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\right) t, \frac{1}{2}\left(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\right) t^{2}$
(b) $\left(\frac{\alpha^{2}-\beta^{2}}{\alpha \beta}\right) t, \frac{1}{2}\left(\frac{\alpha^{2}-\beta^{2}}{\alpha \beta}\right) t^{2}$
(c) $\left(\frac{\alpha+\beta}{\alpha \beta}\right) t, \frac{1}{2}\left(\frac{\alpha+\beta}{\alpha \beta}\right) t^{2}$
(d) $\left(\frac{\alpha \beta}{\alpha+\beta}\right) t, \frac{1}{2}\left(\frac{\alpha \beta}{\alpha+\beta}\right) t^{2}$

Answer 1

Correct Option (d)
Explanation:
Let the car accelerate for time $t_{1}$ travelling distance $S_{1}$ and acquiring maximum velocity $v$.
$$\begin{aligned} &\text { Then, } S_{1}=\frac{1}{2} \alpha t_{1}^{2} \text { [from the equation, } S=u t+\frac{1}{2} a t^{2} \text { ] } \\ &\text { and } \quad \begin{aligned} v &=\alpha t_{1} \\ \Rightarrow \quad & S_{1}=\frac{1}{2} \alpha \times \frac{v^{2}}{\alpha^{2}} \\ &=\frac{v^{2}}{2 \alpha} \end{aligned} \end{aligned}$$
After this car decelerates for time $t_{2}$ to come to rest
Hence, $\quad S_{2}=v t_{2}-\frac{1}{2} \beta t_{2}^{2}$
[from the equation, $S=u t-\frac{1}{2} a t^{2}$ ]
and $0=v-\beta t_{2}$
[from the ecuation, $u=v-a t$ ]
$$\begin{aligned} &\Rightarrow \quad S_{2}=\frac{v^{2}}{\beta}-\frac{v^{2}}{2 \beta}=\frac{v^{2}}{2 \beta} \\ &\text { Now }_{1} \quad \begin{array}{c} t_{1}+t_{2}=t \\ \frac{v}{\alpha}+\frac{v}{\beta}=t \end{array} \\ &\Rightarrow \quad v=\left(\frac{\alpha \beta}{\alpha+\beta}\right) t \end{aligned}$$
Also, total distance travelled,
$$\begin{aligned} S &=S_{1}+S_{2} \\ &=\frac{v^{2}}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \\ &=\frac{1}{2}\left(\frac{\alpha \beta t}{\alpha+\beta}\right)^{2}\left(\frac{\alpha+\beta}{\alpha \beta}\right) \\ &=\frac{1}{2}\left(\frac{\alpha \beta}{\alpha+\beta}\right) t^{2} \end{aligned}$$

Answer 2

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Answer 3

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