A point moves in a straight line so that its displacement x at time t is given by equation x2=t2+1. Its acceleration is:
A
1x
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B
1x3
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C
−1x2
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D
−1x3
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Solution
The correct option is B1x3 Differentiating w.r.t ′t′ on both sides, 2xdxdt=2t+0 xv=t⇒v=tx
Differentiating again w.r.t t, xdvdt+vdxdt=1 xa+v2=1 xa=1−v2 xa=1−t2x2(∵xv=t) xa=1−[x2−1x2](∵x2=t2+1) xa=1−1+1x2 ⇒a=1x3