Question

21. A body moving with uniform acceleration is having
velocity 2 m/s and 8 m/s at t = 1 s and t = 4 s
respectively. Then average velocity of particle in the
time interval t = 1 s to t = 4 s is
(1) 2 m/s
(2) 3 m/s
(3) 5/3 m/s
(4) 5 m/s​

Answer 1

Answer:

a=(v-u) /t

where ,

a= acceleration

v=final velocity

u=initial velocity

t=time

So, the acceleration is (8–2)/(3–1)=2m/s^2

Now,

s= ut+1/2at^2

=(0*4+1/2*2*4^2)m [as the particle started from rest]

=(0+16)m

=16m

For the 1st second displacement was 1/2*2*1^2=1m

Now,

average velocity

=total displacement /total time

=(16–1/4–1)m/s

=5m/s

2 nd method

Area of v-t curve =displacement

So, the displacement from t=1 to t=4sec is the red marked area

Now; to get the red marked area we need to subtract the green marked area from the total area.

[(1/2*4*8)-(1/2*1*2)]m/s*s

=(16–1)m

=15m

average velocity

=total displacement /total time

=(15/3)m/s

=5m/S

Explanation: