Question

When a tiger spots points prey at the distance of $ 200m $ . What is the minimum time it will take to get its prey such that its average velocity $ 90Km{h^{ - 1}} $ ?

Answer 1

Hint :In order to solve this question, we are first going to take the distance and the speed that are given, where the distance is given in meters, speed is in kilometers per hour, so, we convert the speed into meter per second and then calculate the time by dividing the distance and the speed.
To convert the speed in $ Km{h^{ - 1}} $ to $ m{s^{ - 1}} $ , we multiply the speed given, with $ 1000 $ to convert into meters and divide it with $ 3600 $ to convert into second
i.e. $ 1Km{h^{ - 1}} = \dfrac{{1 \times 1000}}{{3600}}m{s^{ - 1}} $
Time taken by the tiger can be calculated as
 $ time = \dfrac{{distance}}{{speed}} $

Complete Step By Step Answer:
In order to solve this question, we are going to first take the distance and the speed and convert them into the S.I. units
As it is given that,
 $ distance = 100m \\
  speed = 90Km{h^{ - 1}} \\ $
Converting $ Km{h^{ - 1}} $ to $ m{s^{ - 1}} $
 $ speed\left( {m{s^{ - 1}}} \right) = \dfrac{{90 \times {{10}^3}}}{{60 \times 60}} = 25m{s^{ - 1}} $
Now we know that the time taken can be calculated by dividing the distance covered with the speed of the tiger
 $ time = \dfrac{{distance}}{{speed}} = \dfrac{{100}}{{25}} = 4s $
Hence, the minimum time it will take to get its prey such that is average velocity is $ 90Km{h^{ - 1}} $ : $ 4s $

Note :
It is important to note that without the conversion of units, a question should never be solved, in this problem, before finding the time, first of all we need to make the distance as well as the speed into a same system of units only, then the time can be found either in hours or in seconds according to the conversion.