A jogger takes 4.5 minutes to run the same distance that a second jogger can run in 3.5 minutes. What is the distance if the second jogger runs 1.5 ft./sec. faster than the first jogger?(Explain)

Answer :

Formula of speed, distance and time is:

[tex]\text{speed}=\frac{dis\tan ce\text{ }}{time}[/tex]

Let distance is:

[tex]x[/tex]

Speed of first jogger is:

[tex]S[/tex]

For first jogger :

[tex]\begin{gathered} \text{speed}=\frac{dis\tan ce}{time} \\ S=\frac{x}{4.5} \end{gathered}[/tex]

For the second jogger distance is same (x) and time is 3.5 minutes and speed is 1.5 then distance is:

[tex]\begin{gathered} \text{time}=3.5\times60 \\ =210\sec \text{.} \end{gathered}[/tex]

[tex]\begin{gathered} \text{speed}=\frac{dis\tan ce}{time} \\ 1.5=\frac{x}{210} \\ x=210\times1.5 \\ x=315\text{ ft} \end{gathered}[/tex]

If distance is 315 ft then speed of first jogger is:

[tex]\begin{gathered} \text{time}=4.5\times60\text{ } \\ =270\text{ sec.} \end{gathered}[/tex]

[tex]\begin{gathered} \text{speed}=\frac{dis\tan ce}{time} \\ S=\frac{x}{270} \\ S=\frac{315}{270} \\ S=1.16\text{ ft./sec.} \end{gathered}[/tex]

Second jogger is faster then first jogger then difference between speed is:

[tex]\text{Faster sp}eed\text{ =spe}ed\text{ of second jogger - sp}eed\text{ of first jogger}[/tex][tex]\begin{gathered} =1.5-1.16 \\ =0.34\text{ ft./sec.} \end{gathered}[/tex]

The second jogger is faster 0.34 ft/sec. then the first jogger.

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