posted by srikala on .
A particle starts moving from rest with uniform acceleration. it travels a distance x in first 2 sec and a distance y in the next 2 sec .Then
X distance moved in first 2 secs.
=> x=0.2+(1/2).a.2^2= 2a ...(1)
velocity at the end of 2 secs.(say v)
=> v = 0+a.2 = 2a ....(2)
y distance covered in the next 2 secs.
=> y = v.t +(1/2).a.t^2
or y = 2a.2 + (1/2).a.2^2 = 6a ...(2)
From, (1) & (2) y = 3x ...Option (2)
a ball dropped on to the floor from a height of 10mrebounds to a height 2.5m. If the ball is in contact with the floor for 0.02s,its avg. acceleration during contact