Question

A tiger running 100 m race, accelerates for one third time of the total time and then moves with uniform speed. Then find the total time taken by the tiger to run 100 m if the acceleration of the tiger is $8m/{s^2}$:

Answer 1

In order to denote the position of an object we define a coordinate system and an origin as a reference point. If a body moves from one position to another position then subtracting the initial position vector from the final position vector gives us displacement. Rate of change of displacement with respect to time gives us velocity. Rate of change of velocity gives us acceleration.

Formula used:
$v = u + at$
$s = ut + \dfrac{1}{2}a{t^2}$

Complete answer:
Rate of change of displacement with respect to time gives us the velocity and the rate of change of velocity with respect to time gives us the acceleration. Now if we are given with acceleration as a function of time and we were asked to find out the displacement then first we should get the velocity function by integrating the acceleration with in the proper limits given and then later after getting the velocity function, integrate it again with respect to time to get the displacement.
Let us assume total time travelled is t. initial velocity(u) is zero. Acceleration(a) upto one third of time is $8m/{s^2}$. Total distance travelled will be 100m.
We will apply one of the kinematic formulas for one third of time to find displacement then.
\eqalign{ & s = ut + \dfrac{1}{2}a{t^2} \cr & \Rightarrow {s_1} = 0 + \dfrac{1}{2}\left( 8 \right){\left( {\dfrac{t}{3}} \right)^2} \cr & \therefore {s_1} = \dfrac{{4{t^2}}}{9} \cr}
For the rest two third of time we have
\eqalign{ & v = u + at \cr & \Rightarrow v = 0 + 8\left( {\dfrac{t}{3}} \right) \cr & \Rightarrow v = \dfrac{t}{3} \cr & \Rightarrow s = ut + \dfrac{1}{2}a{t^2} \cr & \Rightarrow {s_2} = \left( {\dfrac{{8t}}{3}} \right)\left( {\dfrac{{2t}}{3}} \right) + 0 \cr & \therefore {s_2} = \dfrac{{16{t^2}}}{9} \cr}
The sum of two distances will be 100 metres.
\eqalign{ & {s_1} + {s_2} = 100m \cr & \Rightarrow \dfrac{{4{t^2}}}{9} + \dfrac{{16{t^2}}}{9} = 100 \cr & \Rightarrow \dfrac{{20{t^2}}}{9} = 100 \cr & \Rightarrow {t^2} = 45 \cr & \therefore t = \sqrt {45} = 6.708\sec \cr}

The tiger will finish the race in 6.708 seconds.

Note:
In the above solution, first we had found the distance travelled by tiger in one third of the time period as it is in an accelerating phase. After that for two third of the time period it is in a constant speed phase, so we had found that distance again. We had added a total two distances to get a hundred meters.