Question

A hundred-meter sprinter increases her speed from rest uniformly at the rate of $1\;ms^{-2}$ up to three quarters of the total run and covers the last quarter with uniform speed. How much time does she take to cover the first half and the second half of the run?

Answer 1

It is possible to obtain necessary solutions using just the basic kinematic equations of motion. In other words, separating the problem into the first three quarters and the last quarter and then finding the time the sprinter takes to cover them helps.

Also find the velocity that she achieves at the end of the first three quarters since that is the velocity she maintains for the last quarter. And remember that finding the time she takes to cover the first half of the run should account for the acceleration.

Formula used:
Kinematic equations of motion:
$s = ut + \dfrac{1}{2}at^2$ where, $s$ is the distance travelled, $u$ is the initial velocity, $t$ denotes time taken and $a$ is the acceleration.
$v = u + at$, where $v$ is the final velocity, $u$ is the initial velocity, $t$ is the time taken and $a$ is the acceleration.

Complete answer:
Using a simple kinematic equation of motion, we can conveniently deconstruct the scenario into the first three quarters and the last quarter and determine the time the sprinter takes to cover the first and second half of the runs respectively.
Since it is a $100\;m$ sprint, $s_{total} = 100\;m$

Let us first consider the first three quarters:

Here the distance covered will be $s_{3/4} = \dfrac{3}{4} \times 100 \Rightarrow s_{3/4} = 75\;m$
We can obtain the time she takes to run this distance taking $u=0\;ms^{-1}$, $a=1\;ms^{-2}$ and using:
$s_{3/4} = ut_{3/4} + \dfrac{1}{2}at^{2}_{3/4} \Rightarrow t_{3/4} = \sqrt{\dfrac{2 \times s_{3/4}}{a}} \Rightarrow t_{3/4} = \sqrt{2 \times 75} = 5\sqrt{6}\;s$

Following this, the velocity that the sprinter achieves by the end of the third quarter can be determined by:

$v_{3/4} = u + at_{3/4} \Rightarrow v_{3/4} = 0 + 1\times 5\sqrt{6} = 5\sqrt{6}\;ms^{-1}$
This velocity that the sprinter achieves is the same velocity that she maintains through the last quarter. Therefore, considering the last quarter by using $a=0\;ms^{-2}, v_{1/4} = 5\sqrt{6}\;ms^{-1}$ and $s_{1/4} = 25\;m$, we can directly find the time she takes to travel this quarter by:
$t_{1/4} = \dfrac{ s_{1/4}}{ v_{1/4}} = \dfrac{25}{5\sqrt{6}} = \dfrac{5}{\sqrt{6}}\;s$
Therefore the total time she takes to cover the whole sprint is $t_{total} = 5\sqrt{6} + \dfrac{5}{\sqrt{6}} = \dfrac{35}{\sqrt{6}}\;s$
We can now find-

(i)Time to cover first half $t_{first\;half}$
$s_{first\;half} = ut_{first\;half}+ \dfrac{1}{2}at^{2}_{first\;half} \Rightarrow t_{first\;half} = \sqrt{\dfrac{2 \times s_{first\;half}}{a}} = \sqrt{2 \times 50} = 10\;s$

(ii)Time to cover second half$t_{second\;half}$
$t_{second\;half} = t_{total} - t_{first\;half} = \dfrac{35}{\sqrt{6}} -10 = 14.289 – 10 = 4.289\; s$

Note:
Remember that the sprinter accelerates for three quarters of the run. So, while finding the time taken to complete the first half of the run the acceleration is a contributing factor.
It is always easier to find the total time she takes for the whole run since it reduces the work in calculating the time she takes to run the second half, since otherwise, you’ll have to calculate the time she takes to cover the two quarters in the last half separately.