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2 years ago

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A man works on a striaght road from his home to market 2.5km away dwith a speed of 5km/h.Finding the market closed, he instantly

truns and walk back home with a speed of 7.5km/h. The average speed of the man over the interval of time 0.40min.Is equal to?

1 answer:

umka21 [38]2 years ago

7 0

Answer:

5.625km/h

Explanation:

We are given that

Distance between home and market, d=2.5 km

Speed, v1=5km/h

Speed, v2=7.5 km/h

We have to find the average speed of the man over the interval of time 0 to 40min.

Time, $t=\frac{distance}{speed}$

Using the formula

$t_1=\frac{2.5}{5}=0.5 h=0.5\times 60=30 min$

1 hour =60 min

$t_2=\frac{2.5}{7.5}=\frac{1}{3}hour=\frac{60}{3}=20min$

Distance traveled by man in 10 min with speed 7.5 km/h= $\frac{2.5}{20}\times 10=2.5/2km$

Therefore,

Total distance covered=2.5+2.5/2=3.75 km

Time=40 min=40/60=2/3 hour

Average speed= $\frac{total\;distance}{total\;time}$

Average speed= $\frac{3.75}{2/3}=5.625km/h$

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