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A car accelerates from rest with $2 \mathrm{~m} / \mathrm{s}^2$ on a straight line path and then comes to rest after applying brakes. Total distance travelled by the car is $100 \mathrm{~m}$ in 20 seconds. Then, the maximum velocity attained by the car is
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Verified Answer
The correct answer is:
$10 \mathrm{~m} / \mathrm{s}$
Given,
Acceleration of $\operatorname{car}(\alpha)=2 \mathrm{~m} / \mathrm{s}^2$
Distance covered by car in 20 seconds $(s)=100 \mathrm{~m}$
We know that,
$$
\begin{aligned}
s & =\frac{1}{2} \cdot \frac{\alpha \beta}{\alpha+\beta} t^2 \\
100 & =\frac{1}{2} \times \frac{2 \beta}{(2+\beta)} \times 20 \times 20 \\
\frac{100}{400} & =\frac{\beta}{(2+\beta)} \\
\Rightarrow \quad \frac{1}{4} & =\frac{\beta}{2+\beta} \\
\Rightarrow \quad 4 \beta & =2+\beta \\
\Rightarrow \quad 3 \beta & =2 \Rightarrow \text { retardation } \beta=\frac{2}{3} \mathrm{~m} / \mathrm{s}^2 \\
& \therefore \quad \text { Maximum velocity }\left(v_{\text {max }}\right)=\frac{\alpha \beta t}{\alpha+\beta}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{2 \times \frac{2}{3} \times 20}{2+\frac{2}{3}}=\frac{\frac{80}{3}}{\frac{8}{3}}=\frac{80}{3} \times \frac{3}{8} \\
& =10 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Acceleration of $\operatorname{car}(\alpha)=2 \mathrm{~m} / \mathrm{s}^2$
Distance covered by car in 20 seconds $(s)=100 \mathrm{~m}$
We know that,
$$
\begin{aligned}
s & =\frac{1}{2} \cdot \frac{\alpha \beta}{\alpha+\beta} t^2 \\
100 & =\frac{1}{2} \times \frac{2 \beta}{(2+\beta)} \times 20 \times 20 \\
\frac{100}{400} & =\frac{\beta}{(2+\beta)} \\
\Rightarrow \quad \frac{1}{4} & =\frac{\beta}{2+\beta} \\
\Rightarrow \quad 4 \beta & =2+\beta \\
\Rightarrow \quad 3 \beta & =2 \Rightarrow \text { retardation } \beta=\frac{2}{3} \mathrm{~m} / \mathrm{s}^2 \\
& \therefore \quad \text { Maximum velocity }\left(v_{\text {max }}\right)=\frac{\alpha \beta t}{\alpha+\beta}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{2 \times \frac{2}{3} \times 20}{2+\frac{2}{3}}=\frac{\frac{80}{3}}{\frac{8}{3}}=\frac{80}{3} \times \frac{3}{8} \\
& =10 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
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