A point mass starts moving in straight line with constant acceleration a from rest at t=0.At time t=2s ,the acceleration changes the sign,remaining the same in magnitude .The mass returns to the initial position at time t=t0 after start of motion .Here, t0 is=? Ans is (4+2root2)s but how?
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In the 1st case final velocity v is
V= 0+ a*(2) =2a
And distance covered can be calculated by v²-u²=2as
S=4a²/2a =2a
Now in reverse motion the initial velocity will become -2a as the direction is reversed and acceleration will be treated positice(due to change in direction)
Therefore,
For second part of motion
S=(-2a)(t0-t) + 1/2a(t0-t) ²
But S remains the same
Therefore
2a= -2a(t0-t) + a*(t0-t) /2
Therefore on simplifying this equation we get
t0²-8t0+8=0
Thus t0= 4±2root2
But negative sign will not be considered
Hence answer = 4+2root2
Please mark it as the brainliest answer
V= 0+ a*(2) =2a
And distance covered can be calculated by v²-u²=2as
S=4a²/2a =2a
Now in reverse motion the initial velocity will become -2a as the direction is reversed and acceleration will be treated positice(due to change in direction)
Therefore,
For second part of motion
S=(-2a)(t0-t) + 1/2a(t0-t) ²
But S remains the same
Therefore
2a= -2a(t0-t) + a*(t0-t) /2
Therefore on simplifying this equation we get
t0²-8t0+8=0
Thus t0= 4±2root2
But negative sign will not be considered
Hence answer = 4+2root2
Please mark it as the brainliest answer
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