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A man is on a journey on a straight road from his home to a market 2.5km away with a speed of 5kmh1. It has been found that when the market closes, he instantly turns and walks back home with a speed of 7.5kmh1.
(a) What will be the magnitude of average velocity?
(b) What will be the average speed of the man over the interval of time?
(1)0 to 30min(2)0 to 50min(3)0 to 40min

Answer
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Hint: First of all find the average velocity and average speed on the journey to the market and the comeback. Calculate the total time taken for the travel. When the person comes back home, the displacement will be zero. Calculate the same in each interval of time mentioned in the question. This all will help you in answering the question.

Complete step by step answer:
The time taken to reach market will be,
t1=2.55=0.5h=30min
The time taken to get back to home will be calculated as,
t2=2.57.5=0.33h=20min
The average velocity for the interval (030)min will be,
v=2.50.5=5kmh1
Average speed for the interval (030)min will be,
s=2.50.5=5kmh1
Total time taken for the travel will be,
 t=30+20=50min=56h
We can see that when he reached back then net displacement will be zero
Therefore in the case of (050)min, the average velocity will be,
v=056=0kmh1
Total distance he covered when he arrive back will be,
2.5+2.5=5km
Therefore the average speed will be given as,
s=556=6kmh1
Distance travelled in the first 30minis given as 2.5km. While returning, the distance travelled in 10min,
d=7.5×16=1.25km
The total time taken will be,
t=40min=23h
Hence the displacement in 040minwill be,
2.51.25=1.25km
The average velocity for the interval 040minis,
v=1.2523=1.875kmh1
The average speed in this interval is given as,
s=3.7523=5.625kmh1
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Note: The velocity is the time rate of variation of the displacement. Distance covered is the total length of the path traversed by a body. The displacement will be the shortest distance between the initial and final locations of the body. Time rate of variation of distance is known as speed.