Question

A police inspector in a jeep is chasing a pickpocket on a straight road. The jeep is at maximum speed "v"(assume uniform). The pickpocket rides on the motorcycle of a waiting friend when the jeep is at a distance "d" away and the motorcycle starts with a constant acceleration "a". Show that the pickpocket will be caught if:
$\sf{v \geqslant \sqrt{2ad} }$

Detailed explanation required

Class 11

2D Kinematics

​

Answer 1

Answer:

see the attachment for the answer

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Jeeps maximum speed = v

Distance between them = d.

Acceleration of motor bike = a.

Initial velocity of motor bike = 0 m/s.

$\huge{\bold{\underline{Explanation:-}}}$

Suppose the thief is caught at time = "t".

then,

Applying Second kinematic equation.

$\large{\bold{S = ut + \frac{1}{2}at^2}}$

Substituting the values.

$\large{S = \frac{1}{2}at^2 -----(1)}$

The "S" is the distance travelled by the thief before police catches the thief.

As mentioned in question Speed of jeep is uniform which means acceleration is Zero.

Again, applying second kinematic equation.

$\large{\bold{S' = ut + \frac{1}{2}at^2}}$

Substituting the values.

When the police catches the thief it travels a distance of S' = S + d

$\large{S + d = vt + 0}$

$\large{S + d = vt }$

Arranging.

$\large{S = vt - d ----(2)}$

Compare equation (1) and(2)

$\large{ \frac{1}{2} at^2 = vt - d}$

$\large{ at^2 = 2vt - 2d}$

$\large{at^2 - 2vt + 2d = 0}$

We need to find velocity but if we simply this we will get time.

So, Time will be real when police catches the thief

Then We only need to find discriminant.

so,Discriminant = 0

$\large{\bold{b^2 - 4ac \geqslant 0}}$

Substituting the values.

$\large{(2v)^2 - 4 \times a \times 2d \geqslant 0 }$

$\large{ \cancel{4}v^2 \geqslant \cancel{4} \times 2ad}$

Now,

$\large{v^2 \geqslant 2ad}$

$\huge{\boxed{\boxed{ v \geqslant \sqrt{2ad}}}}$

Hence derived.