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Question

A particle moves along the x-axis. Its velocity time-graph is given below.

The particle changes its direction of acceleration, when

A
t<1s
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B
t=2s
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C
t>1s
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D
t1 s
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Solution

The correct option is C t>1s
As the curve is parabolic let us assume a second order equation for velocity to fit in this curve
Let v=at2+bt+c(1)
Apply boundary condition to above equation to find a,b and c
At t=0 s, v=0 ms2
0=(a×02)+(b×0)+c
c=0
At t=1 s, v=3 ms2
3=(a×12)+(b×1)+0
3=a+b(2)
At t=2 s, v=0 ms2
0=(a×22)+(b×2)
b=2a(3)
Substitute (3) in equation (2) we get,
3=a2a
a=3
Substitute value of a in equation (2)
3=3+b
b=6
Substitute these a, b and c in equation (1)
v=3t26t
dvdt=6t6
For t1, dvdt0
For t>1, dvdt>0
Acceleration changing its direction after 1 s
t>1s
Acceleration at any point is the slope of vt graph at that point. From t=0 to 1 s, slope is negative but after t=1 s, slope of vt graph becomes positive, which means direction of acceleration changes after t=1 s.

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