Question

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. Which of the following relations is true if the police are able to catch the dacoit?
A. ${{v}^{2}}\le \alpha x$
B. ${{v}^{2}}\le 2\alpha x$
C. ${{v}^{2}}\ge 2\alpha x$
D. ${{v}^{2}}\ge \alpha x$

Answer 1

We should consider a point which is located at a distance S from the police and s from the dacoit. Time taken by both to cover the distances is the same, t. Also, we are given in the question that dacoit located at distance x from jeep has an initial velocity zero (rest) and constant acceleration ∝, from which we could find s with the help of Newton’s equation of motion. Equating the sum of x and s to S gives you a quadratic equation in t. You could then solve for a real value of t.
Formula used:
Newton’s Equation of motion,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
Expression for distance,
$S=v\times t$

Complete step by step answer:
We are given that the police are chasing the dacoit in a jeep moving at constant speed v.
Also, the dacoit accelerates his motorcycle at a constant acceleration ‘∝’ from rest when he is at a distance ‘x’ from the jeep.
We are asked to find the condition for which the police catch the dacoit.
Let us consider a point A, such that the police catch the dacoit after a time t at this point.
From the equations of motion we have,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ …………… (1)
Where, s = distance travelled
u = initial velocity
a= acceleration of motion
t = time taken
For the case of dacoit’s motion, he covers the distance from B to A with zero initial velocity, acceleration and time t. So, equation (1) for the dacoit can be given by,
$s=\dfrac{1}{2}\alpha {{t}^{2}}$ ………………. (2)
Police jeep has to cover a distance x along with the distance covered by dacoit in order to catch the dacoit at point A. So the distance travelled by the police jeep is given by,
$S=x+s$ …………………. (3)
We know that distance= velocity × time taken. That is,
$S=v\times t$ ………………. (4)
Since, the time taken by both police and dacoit is the same we represent both time by t.
Substituting (4) and (2) in (3), we get,
$v\times t=x+\dfrac{1}{2}\alpha {{t}^{2}}$
$\alpha {{t}^{2}}-2vt+2x=0$
Now let us solve this quadratic equation in t.
$t=\dfrac{2v\pm \sqrt{4{{v}^{2}}-8\alpha x}}{2}$
We are to find a value for t such that it is a real number. So,
$4{{v}^{2}}-8\alpha x\ge 0$
$4{{v}^{2}}\ge 8\alpha x$
${{v}^{2}}\ge 2\alpha x$
Hence, we get the answer to the given question as option C.

Note:
Since the time has to be a real number, we are to solve the quadratic equation in t for a real number. For that, the discriminant has to be positive. The discriminant is the term that comes inside the root while solving a quadratic equation, that is, ${{b}^{2}}-4ac$ for the equation $a{{x}^{2}}+bx+c=0$.