Who will win?

Classical Mechanics Level 3

Two cars $A$ and $B$ are racing along a straight line, car $A$ is leading and their relative velocity is directly proportional to the distance between the two cars. When the lead of car $A$ is $x_{1}=10\text{ m}$ , it is running $10\text{ ms}^{-1}$ faster than car $B$ . If the time taken by car $A$ to increase its lead to $x_{2}=20\text{ m}$ from car $B$ is in the form $t=\ln n$ , find $n$ .

The answer is 2.

1 solution

Akshat Sharda
Jun 15, 2016

It is given that $v_{\text{rel}}\propto x$ where $v_{\text{rel}}$ is the relative velocity and $x$ is the distance between the two cars.

$\therefore v_{\text{rel}}=kx \\ 10=kx_{1}=10k\Rightarrow k=1 \\ \therefore v_{\text{rel}}=x \Rightarrow \frac{dx}{dt}=x \Rightarrow \frac{dx}{x}=dt \\ \int_{10}^{20} \dfrac{dx}{x}=\int_{0}^{t} dt \Rightarrow \ln x |_{10}^{20}=t|_{0}^{t} \\ t=\ln{20} -\ln{10}=\ln 2 \\ \Rightarrow \boxed{2}$

Who will win?

The answer is 2.

1 solution

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