A car accelerates from rest at a constant rate
Question:

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is $t$ seconds, the total distance travelled is:

  1. (1) $\frac{4 \alpha \beta}{(\alpha+\beta)} t^{2}$

  2. (2) $\frac{2 \alpha \beta}{(\alpha+\beta)} \mathrm{t}^{2}$

  3. (3) $\frac{\alpha \beta}{2(\alpha+\beta)} \mathrm{t}^{2}$

  4. (4) $\frac{\alpha \beta}{4(\alpha+\beta)} \mathrm{t}^{2}$


Correct Option: , 3

Solution:

(3)

$\mathrm{v}_{0}=\alpha \mathrm{t}_{1}$ and $0=\mathrm{v}_{0}-\beta \mathrm{t}_{2} \Rightarrow \mathrm{v}_{0}=\beta \mathrm{t}_{2}$

$\mathrm{t}_{1}+\mathrm{t}_{2}=\mathrm{t}$

$\mathrm{v}_{0}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\mathrm{t}$

$\Rightarrow \mathrm{v}_{0}=\frac{\alpha \beta t}{\alpha+\beta}$

Distance $=$ area of $\mathbf{v}-\mathrm{t}$ graph

$=\frac{1}{2} \times \mathrm{t} \times \mathrm{v}_{0}=\frac{1}{2} \times \mathrm{t} \times \frac{\alpha \beta \mathrm{t}}{\alpha+\beta}=\frac{\alpha \beta \mathrm{t}^{2}}{2(\alpha+\beta)}$

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