Question

A thief is running away on a straight road on a jeep moving with a speed of 9m/s. A policeman chases him on a motor cycle moving at a speed of 10m/s.If the instantaneous separation of jeep from the motor cycle is 100m,how long will it take for the policemen to catch the bus?

Answer 1

Answer:

Given:

Policeman's velocity = 10 m/s

Thief's velocity = 9 m/s

Instantaneous distance = 100 m

To find:

Time taken to catch the thief.

Calculation:

Relative velocity between the policeman and the thief

= V(policeman) - V"(thief)

= 10 - 9

= 1 m/s

Since velocity of policeman is greater than that of thief, the policeman will surely catch the thief.

Time taken =

(Instantaneous distance)/(Rel. velocity)

= 100/1

= 100 seconds.

So the final answer is 100 seconds.

Answer 2

Answer:

• Policemen will take 100 seconds to catch the Theif.

Explanation:

$\rule{300}{1.5}$

Relative velocity:-

• It's the Difference between two velocities of two Different bodies.

We know,

$\large{\boxed{\bold{v_r = v_2 - v_1}}}$

$\large\bold{Given}\begin{cases}\sf{v_2 = Velocity \: of \: Policemen = 10 \: m/s} \\ \sf{v_1 = Velocity \: of \: Thief = 9 \: m/s}\end{cases}$

Instantaneous Time:-

• It is the Ratio of Instantaneous Distance by Velocity.

We know,

$\large{\boxed{\bold{T = \dfrac{D}{v_r}}}}$

$\large \bold{Given}\begin{cases}\sf{v_r = Relative \: Velocity } \\ \sf{D = Instantaneous \: separation = 100 \:m}\end{cases}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Relative velocity,

$\large{\boxed{\bold{v_r = v_2 - v_1}}}$

Substituting the values,

$\large{\tt \hookrightarrow v_r = 10 \: m/s - 9 \: m/s}$

$\large{\tt \hookrightarrow v_r = 10 - 9}$

$\large{\boxed{\tt v_r = 1 \; m/s}}$

From Instantaneous Time,

$\large{\boxed{\bold{T = \dfrac{D}{v_r}}}}$

$\large{\tt \hookrightarrow T = \dfrac{100 \: m}{1 \: m/s}}$

$\large{\tt \hookrightarrow T = \dfrac{100}{1}}$

$\huge{\boxed{\boxed{\tt T = 100 \: Sec}}}$

So, The Policeman will catch the thief (or) Jeep in 100 seconds.

$\rule{300}{1.5}$