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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.

Answer and Explanation:

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.


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