Question

Inraces, runner 1 on track 1 (with time $\mathbf{2}\mathbf{}\mathbf{min}\mathbf{,}\mathbf{}\mathbf{27}\mathbf{.}\mathbf{95}\mathbf{}\mathbf{sec}$) appears to be faster than runner 2 on track 2 ($\mathbf{2}\mathbf{}\mathbf{min}\mathbf{,}\mathbf{}\mathbf{28}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{sec}$). However, length $L_2$of track 2 might be slightly greater than length $L_1$of track 1. How large $L_2-L_1$can be for us still to conclude that runner 1 is faster?

Short answer

The value of to conclude that runner 1 is faster is less than 1.35 m .

Step-by-step solution

Given data

Length of track 1,$L_1=1\mathrm{km}=1000\mathrm{m}$,

Time taken by runner 1,$t_1=2\min ,27.95\sec$

Time taken by runner 2, $t_2=2\min ,28.15\sec$

 Step 2: Understanding the speed

Speed may be defined as the rate of change of distance with time. The speed of runner 1, when multiplied by the time taken by runner 2, gives the actual length of track 2 to conclude that runner 1 is faster.

The equation for the speed is given as follows:

$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$

Determination of the value of L2-L1  for which runner 1 will be faster

Time taken by runner 1 is calculated as follows:

$\begin{array}{c}{t}_1=2\times 60+27.95\\ =147.95\mathrm{s}\end{array}$

Time taken by runner 2 is calculated as follows:

$\begin{array}{c}{t}_2=2\times 60+28.15\\ =148.15\mathrm{s}\end{array}$

The speed of runner 1 is calculated as follows:

$\begin{array}{c}{v}_1=\frac{L_1}{t_1}\\ =\frac{1000\mathrm{m}}{147.95\mathrm{s}}\\ =6.759\mathrm{m}/\mathrm{s}\end{array}$

Now, the distance covered by runner 2 with this speed is calculated as follows:

role="math" localid="1651842941901" $\begin{array}{c}{L}_2=v_1\times t_2\\ =6.759\mathrm{m}/\mathrm{s}\times 148.15\mathrm{s}\\ =1001.35\mathrm{m}\end{array}$

So, the value of $L_2-L_1$is as follows:

$\begin{array}{c}{L}_2-L_1=1001.35\mathrm{m}-1000\mathrm{m}\\ =1.35\mathrm{m}\end{array}$

Thus, the value of $L_2-L_1$to conclude that runner 1 is faster is less than 1.35 m