Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A particle moves as such acceleration is given by\[\alpha =3\text{ }sin\text{ }4t,\] then: (1) the initial velocity of the particle must be zero (2) the acceleration of the particle becomes zero after each interval of \[\frac{\pi }{4}\] second (3) the particle comes at its initial position after sometime (4) the particle must move on a circular pathA) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 are correct
D) 1, 2 and 3 are correct
Correct Answer: B
Solution :
p(1) \[\therefore \] \[a=3\sin 4t\] \[\frac{dv}{dt}=3\sin 4t\] \[\therefore \] \[\int{dv}=\int{3\sin 4t+c}\] \[\therefore \] \[v=-\frac{3}{4}\cos 4t+c\] where c is constant of integration At \[t=0,{{v}_{o}}=-\frac{3}{4}+c\] (initial velocity) If \[c=\frac{3}{4},{{v}_{o}}=0\] Hence initial velocity may or may not be zero. (2) Since, \[a=3\text{ }sin\text{ }4t\] So, \[3\text{ }sin\text{ }4t=0\] (for zero acceleration) \[\therefore \] \[sin\text{ }4t=0\] \[\Rightarrow \] \[4t=n\pi \] \[\Rightarrow \] \[t=\frac{n\pi }{4}\] \[n=0,1,2....\] Hence, the acceleration of the particle becomes zero after each interval of\[\frac{\pi }{4}\]second. (3)Since acceleration is sine function of time, so particle repeats its path periodically. Due to this paritcle comes at its initial position periodically. (4) The path of particle is a straight line.
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