Each day Kim jogs from her home to the lake and then walks back again along the same route. Her average speed is 5m/s (meters per second) while jogging and 2 m/s while walking. If the whole trip takes 35 minutes, how far is it from Kim’s home to the lake?

Respuesta :

Answer:

The distance from Kim’s home to the lake is 3000 meters.

Explanation:

Since the whole trip takes 35 minutes, and the speeds are in meter/second, first convert 35 minutes to seconds.

[tex]35\text{ minutes}=35\times60=2100\text{ seconds}[/tex][tex]\begin{gathered} Average\; \text{Speed}=\frac{\text{Distance}}{Time} \\ \implies\text{Dis}\tan ce=Average\; \text{Speed}\times Time \end{gathered}[/tex]

• Her average speed while jogging = 5m/s

,

• Her average speed while walking = 2m/s

• If the number of seconds spent walking = x seconds.

,

• The number of seconds spent jogging = (2100-x) seconds.

• Distance covered walking = 2x

,

• Distance covered jogging = 5(2100-x)

Since the distance walked and jogged is the same, we have that:

[tex]2x=5(2100-x)[/tex]

We solve for x.

[tex]\begin{gathered} 2x=5(2100-x) \\ 2x=10500-5x \\ 2x+5x=10500 \\ 7x=10500 \\ x=\frac{10500}{7} \\ x=1500\text{ seconds} \end{gathered}[/tex]

Substitute x=1500 to find the required distance.

[tex]\text{Distance covered walking=}2x=2\times1500=3000\text{ meters}[/tex]

The distance from Kim’s home to the lake is 3000 meters.