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A particle moves in East with velocity 15 m per second for 2 seconds. Then moves northwards with 5m per second for 8 seconds . Then average velocity of the particle is ?

A particle moves in East with velocity 15 m per second for 2 seconds. Then moves northwards with 5m per second for 8 seconds . Then average velocity of the particle is ?

Grade:11

5 Answers

Sanskar sharma 814
42 Points
6 years ago
Distance covered by particle in East is - 15×2=30mDistance covered in North is5×8=40m Therefore net displacement is 50m and total time is 10 sec therefore average velocity is 50÷10= 5m/s
Harshit Panwar
108 Points
6 years ago
distance travelled in east=15*2=30m
distance travelled in north=5*8=40m
since east and north are perpendicular to each other.
So, displacement=sq. root of [(30)2+(40)2]=50m
average velocity=dispacement/total time taken=50/(8+2)=5m/s
Mahak
11 Points
6 years ago
Distance covered in east direction=15×2=30m. Distance covered in northward direction=5×8=40m. Average speed=total distance÷total time taken=30+40÷2+8=70÷10=7ms^2
Aawadhut prakashrao Tekale
16 Points
5 years ago
a particle move toward east 15m/s
for 2s =1s=15m then,
           2s=15×2=30m/s.........(1)
then,
   particle move toward north 5m/s for
   8s=8×5=40m/s.............(2)
 average velocity=total displacement/time
        by hypotenous=displacement
so
50m/s=diplacement;
time=2+8=10;
ave.velo=50m/10s
=5m/s
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Distance covered by particle in East is - 15×2=30mDistance covered in North is5×8=40m Therefore net displacement is 50m and total time is 10 sec therefore average velocity is 50÷10= 5m/s

Thanks and Regards

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