Question

A boy is running on a straight road.He runs 500m towards north in 2min 10sec and then turns back and runs 200m in 1 minute.Calculate his average speed and magnitude of average velocity during first 2min 10sec and his average speed and magnitude of average velocity during the whole journey.

Answer 1

For first half of journey till t=2mins10secs=130secs,
distance=displacement=500m
speed=$\frac{Total distance travelled}{total time taken}$
speed=$\frac{500}{130}$
          =3.84m/sec       
velocity=$\frac{Total displacement}{total time taken}$
 velocity=$\frac{500}{130}$  
            =3.84m/sec  towards north
For whole journey, when t=3mins 10 secs= 190secs
distance=(500+200)m=700m
displacement=(500-200)m=300m towards north
speed=$\frac{Total distance travelled}{total time taken}$
          =$\frac{700}{190}$
          =3.78m/sec
velocity=$\frac{Total displacement}{total time taken}$
            =$\frac{300}{190}$
            =1.57m/sec towards noth

Answer 2

For first half of journey till t=2mins10secs=130secs,

distance=displacement=500m

speed=

speed=

         =3.84m/sec       

velocity=

 velocity=  

           =3.84m/sec  towards north

For whole journey, when t=3mins 10 secs= 190secs

distance=(500+200)m=700m

displacement=(500-200)m=300m towards north

speed=

         =

         =3.78m/sec

velocity=

           =

           =1.57m/sec towards noth