Question

A man walks on a straight road from his home to market 2.5kn away with a speed of 5km/hr . finding the market closed he instantly walks back home with a speed of 7.5km/hr . find the magnitude for avg speed and avg velocity over the following time
(a) 0 to 30 minutes.
(b) 0 to 50 minutes.
(c) 0 to 40 minutes.

( With each and every step if possible. )

Answer 1

@anusha.p can you help me with this sum .. please ?

Answer 2

i'm very bad at physics..
i'm sorry...@ashna

Answer 3

its okay .. btw tnx :)

Answer 4

avg speed= total distance covered/ time taken
avg velocity= total displacement/time taken

a) In the first half hour, the man covers 2.5 km(his speed is 5km/h). So the avg speed is 5km/h(given). Or use the above formula. Distance = 2.5km. Time is 1/2h
The avg velocity is also the same since distance and displacement are equal.

b) For the first 50 min
-For the first 30 min
distance covered is 2.5 km
-For the next 20 min(1/3 hours), the man covers a distance of 7.5*1/3=2.5 km.
So the total distance covered= 2.5+2.5
time taken=50 min=5/6 hrs
Avg speed=6km/h

Here the avg velocity will be different since distance and displacement are not equal.
Here,the distance covered in the last 20min at 7.5km/h will get subtracted from the distance covered in the first 30min to get displacement(which is 0).So the avg velocity is 0.