The acceleration-time graph of a particle moving
a straight line along x-axis is given below. If init
velocity of the particle is 4 m/s, then velocity
t = 2 s is
pa (m/s)
sk
T
23
(s)
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Answer:
area of graph ABCD = velocity change
area ABCD = Vf - Vi
BC/AD =BE/AE
BC/2 =1/3
BC =2/3
area of ABCD = Vf -Vi
1/2(2+2/3)×2 =Vf - 4
8/3=Vf - 4
Vf = 8/3 +4
Vf = 8+12/3
Vf =20/3
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