If motion of a particle is represented by s=et+4t3−cost, then the acceleration of the particle at any time t is
[Hint: Acceleration of the particle is d2sdt2]
A
et+12t2+sint
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B
et+24t−cost
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C
et+24t+cost
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D
et+12t2−sint
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Solution
The correct option is Cet+24t+cost As we know that the acceleration of the particle at any time t is given by d2sdt2⇒d(dsdt)dt
Given, S=et+4t3−cost
So, we have dsdt=d(et+4t3−cost)dt ⇒detdt+4dt3dt−dcostdt=et+12t2+sint
Thus, d2sdt2=d(et+12t2+sint)dt ⇒detdt+12dt2dt+dsintdt=et+24t+cost