Question
A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t3−2t), where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin
Answer: Option B
:
B
Given v = 4t3−2t .
∴a=dvdt=12t2−2
andx=∫vdt=∫(4t3−2t)dt=t4−t2
When particle is at 2m from the origin, we have
t4−t2=2
⇒t4−t2−2=0
(t2−2)(t2+1)=0⇒t=√2sec (only possible physical root)
Acceleration at t=√2 sec given by,
a=12t2−2=12×2−2=22m/s2
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:
B
Given v = 4t3−2t .
∴a=dvdt=12t2−2
andx=∫vdt=∫(4t3−2t)dt=t4−t2
When particle is at 2m from the origin, we have
t4−t2=2
⇒t4−t2−2=0
(t2−2)(t2+1)=0⇒t=√2sec (only possible physical root)
Acceleration at t=√2 sec given by,
a=12t2−2=12×2−2=22m/s2
Was this answer helpful ?
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