# A runner is planning for a 10km race. She can maintain a steady speed of 4.1m/s for as much time...

## Question:

A runner is planning for a 10km race. She can maintain a steady speed of 4.1m/s for as much time as needed before ending the race with a 7.8m/s sprint. If she wants to finish in 40 min or less, how far from the finish should she begin to sprint?

## Solving of Degree One Polynomials in Two Variables

The most general form of a first degree polynomial equation in two variables is {eq}a x + b y + c = 0 {/eq}. Here a , b, and c are the constants and x , y are the variables. Such degree one polynomials are known as linear polynomials. In order to solve a first degree polynomial equation in two variables we need a minimum of two equations relating the variables. So from the given situation we need to frame two equations in terms of the variables chosen.

Given points

• Distance to be covered in the race s = 10,000 m
• The steady speed the runner can maintain for long time {eq}v_1 = 4.1 \ m/s {/eq}
• The sprinting speed possible for the runner {eq}v_2 = 7.8 \ m/s {/eq}
• The runner wants to finish the race in time duration t = 40 minutes = 2400 sec.

Let {eq}t_1, \ \ t_2 {/eq} be the time duration for which the runner runs at medium speed and sprints respectively.

Then we have the equation {eq}t_1 + t_2 = 2400 \ s {/eq}

So the time duration for sprinting {eq}t_2 = 2400 - t_1 {/eq}

So we can write the distance covered by the runner in the given intervals of time as {eq}v_1 t_1 + v_2 t_2 = 10,000 {/eq}

Substituting the expression for sprinting time in the equation we get

{eq}v_1 t_1 + v_2 \times ( 2400 - t_1 ) = 10000 {/eq}

So the time duration for which the sprinter may run at medium speed {eq}t_1 = \dfrac { 10000 - v_2 \times 2400 } { v_1 - v_2 } \\ t_1 = \dfrac { 10000 - 7.8 \times 2400 } { 4.1 - 7.8 } \\ t_1 = 2356.757 \ \ s {/eq}

So the time duration for which the runner has to sprint {eq}t_2 = 2400 - t_1 \\ t_2 = 2400 - 2356.757 \\ t_2 = 43.243 \ s {/eq}

Distance covered by the runner in sprinting {eq}s_2 = v_2 t_2 \\ s_2 = 7.8 \times 43.243 \\ s_2 = 337.295 \ m {/eq}

So the runner must start sprinting from a distance of 337.295 meter from the finishing point.