A thief is running away on a straight road with a speed of $9m{s^{ - 1}}$. A policeman chases him in a jeep moving at speed of $10m{s^{ - 1}}$ . If the instantaneous separation of the jeep from the motorcycle is 100m, how long will it take for the police man to catch the thief?
A $1s$
B $19s$
C $90s$
D $100s$

Hint: Speed is one of the basic concepts in physics, which generally explains all about how fast or slow any object is moving. Speed is inversely proportional to the time taken when the distance travelled is constant. So, in other words we can say that whenever the speed increases then the time decreases and vice versa. Speed is a scalar quantity which generally depends on magnitude and independent of direction of motion.


Complete Solution step by step:
Speed: The rate at which an object is moving is known as its speed. It is distance travelled per unit time. The SI unit of distance is $m{s^{ - 1}}.$
Distance: It is defined as the total path travelled by the moving object. The SI unit of distance is meter \[\left( m \right)\].
Speed of thief\[ = 9m{s^{( - 1)}}\].
Speed of police$ = 10m{s^{ - 1}}$
Instantaneous separation (distance) between jeep and motorcycle\[ = 100m\].
We are comparing the speed of thief with police
Hence, the relative speed between them is given by
${v_{relative}} = {v_{police}} - {v_{thief}}$
${v_{relative\,}} = 10 - 9$
${v_{relative}} = 1m/s$
As we know that,
$Speed = \dfrac{{Distance}}{{Time}}$
For time, $Time = \dfrac{{Distance}}{{Speed}}$
Since relative speed is,
${v_{relative}} = 1m/s$
On putting the value of distance and relative speed in above equation,
We get, $Time = \dfrac{{100}}{1}$
So, time taken by police to catch the thief \[ = 100s.\]
Hence, the correct option is D.

Note: The problem was based on Relative motion in a straight line, when the comparison between the moving object takes place then the relative velocity of one with respect to other is needed. When the moving bodies are in the same direction then subtract one with another and for the opposite direction it will be added.