A point moves in a straight line so that its displacement x metre at a time t second is such that t=(x2−1)1/2. Its acceleration in m/s2 at time t second is:
A
1x−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1x3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
−tx2−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
tx2−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B1x3
Given,
t=(x2−1)1/2
Squaring on both sides
t2=x2−1
⇒x2=t2+1
Differentiating with respect to t
2xdxdt=2t as dxdt=v
v=tx
Differentiating both sides (w.r.t to t), we will get,