Question

A boy runs on a straight road from his home to a market 2.0 km away with a speed of 4 kmh.
Finding the market closed, he at once runs back to his home with a speed of 6 km h. Calcualte his
(1) average speed (ii) magnitude of average velocity and (iii) average speed of the boy in the time
interval n(a) 0 to 30 minutes. (b) 0 to 40 minutes.​

Answer 1

1)avg.speed=total dist / total time

Total distance=2+2=4km

Total time =time home to market +time market to home

=2/4+2/6

0.83

Avg speed =4/0.83=4.8kmperhour

2)avg vel=0 since total displacement is 0

3)in 30 min the boy travelled 2km =4kmperhour

In40 min he travelled 2+1=3km

Avg speed=3/1.5=2kmperhour

Answer 2

Answer:

(i) 4.8 km/hr

(ii) Zero (0)

(iii)  (a) 4km/hr

     (b) 4.5km/hr

Explanation:

We know that:

Time= Distance/Speed

So, t1 = 2/4= 0.5hr= 1/2 hr

     t2 = 2/6= 1/3 hr

T= t1 + t2= 1/2 + 1/3= 5/6 hrs

(i) Ave. Speed= Total Distance/Total time taken= 2+2/5/6= 4*6/5= 4.8hrs

(ii) Ave Velocity= Total Displacement/Total time taken= 0/5/6= 0

(since total displacement=0)

(iii) (a) 0 to 30mins ⇒ Onward journey

Ave. Speed= 4km/hr

     (b) 0 to 40mins

time= 10 mins= 10/60= 1/6 hr

speed= 6km/hr

distance= speed * time= 6*1/6= 1km

Ave. Speed= Total Distance/Total time taken

where, total distance= 2+1= 3km

total time taken= 30 + 10= 40mins= 40/60= 2/3 hrs

Therefore,

                 Ave. Speed= 3/2/3= 3* 3/2= 4.5 km/hr

Thank you