Question Details

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. The average speed of the man over the interval of time 0 to 40 min. is equal to

Options

A

5 km/h

B

25/4 km/h

C

30/4 km/h

D

45/8 km/h

Correct Answer :

45/8 km/h

Solution :

The correct answer is Option 4: 45/8 km/h.

Let's break down the motion of the man step-by-step to find his average speed over the time interval from 0 to 40 minutes.

Step 1: Analyze the journey from home to the market
The distance from home to the market is:
d=2.5 km
The speed of the man while walking to the market is:
v1=5 km/h
The time taken to reach the market (t1) is given by the formula:
t1=DistanceSpeed=2.55=0.5 hours
Converting this time into minutes:
t1=0.5×60 min=30 min

Step 2: Analyze the return journey from the market back towards home
The total time interval we are interested in is 40 minutes (from t=0 to t=40 min).
Since the man spent the first 30 minutes walking to the market, the remaining time for the return journey is:
t2=40 min-30 min=10 min
Converting this remaining time into hours:
t2=1060 hours=16 hours
During this return journey, the man walks back with a speed of:
v2=7.5 km/h=152 km/h
The distance covered by the man in these 10 minutes (d2) is:
d2=v2×t2=7.5×16=7.56=1.25 km

Step 3: Calculate the total distance covered and total time
The total distance covered by the man in the 40-minute interval is:
dtotal=d1+d2=2.5 km+1.25 km=3.75 km
Expressing 3.75 as a fraction:
dtotal=154 km
The total time interval is:
ttotal=40 min=4060 hours=23 hours

Step 4: Calculate the average speed
Average speed is defined as the ratio of the total distance covered to the total time taken:
vavg=dtotalttotal
Substituting the values we obtained:
vavg=15423=154×32=458 km/h

Thus, the average speed of the man over the interval of 0 to 40 minutes is 45/8 km/h.

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