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# A police party is chasing a dacoit in a jeep which is moving at a constant speed v. The dacoit is on a motorcycle. When he is at a distance x from the jeep, he accelerates from rest at a constant rate$\alpha$. Which of the following relations is true, if the police is able to catch the dacoit?(a) ${v^2} \leqslant \alpha x$ (b ${v^2} \leqslant 2\alpha x$ (c) ${v^2} \geqslant 2\alpha x$ (d) ${v^2} \geqslant \alpha x$

Last updated date: 20th Jun 2024
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Hint: The above problem can be solved by using the principle of kinematics. The police would catch the dacoit if the police cover the distance that is equal to the distance covered by the dacoit plus the initial distance between the car and jeep in the same duration.

Given: The speed of the dacoit is v, the initial distance between the jeep and motorcycle is x, the acceleration of the motorcycle is $\alpha$.
The distance covered by the dacoit on the motorcycle is given as:
$d = vt......\left( 1 \right)$
The distance covered by the jeep to catch the dacoit is given as:
$X = \dfrac{1}{2}a{t^2} + x......\left( 2 \right)$
Equate the equation (1) and equation (2) to find the required relation.
$X = d$
$\dfrac{1}{2}\alpha {t^2} + x = vt$
$\alpha {t^2} + 2x = 2vt$
$\alpha {t^2} - 2vt + 2x = 0......\left( 3 \right)$
The police catch the dacoit if the roots of the quadratic equation (3) are real and unequal. The discriminant of the quadratic equation for real and unequal roots is given as:
$D \geqslant 0$
The discriminant of the quadratic equation (3) is given as:
${\left( { - 2v} \right)^2} - 4\left( \alpha \right)\left( {2x} \right) \geqslant 0$
${v^2} - 2\alpha x \geqslant 0$
${v^2} \geqslant 2\alpha x$

Thus, the true relation for catching the dacoit is ${v^2} \geqslant 2\alpha x$ and the option (c) is the correct answer.

Note: The above problem can also be solved by using the concept of the relative motion. The dacoit can be assumed stationary at some separation and police moves relative to the dacoit.