Question

For one dimensional motion of a particle, described by x=t-sint​

Answer 1

Explanation:

The position of the particle is given as a function of time.

x=t−sint 

The velocity can be obtained by differentiating the given expression. 

v=dtdx=dtd[t−sint]=1−cost

The acceleration can be obtained by differentiating the expression of velocity w.r.t. time

a=dtdv

a=dtd[1−cost]=sint

As acceleration a>0 for all t>0

Hence, x(t)>0 for all t>0

velocity v=1−cost

if, cost=1,

the velocity will be v=0

vmax=1−(cost)min=1−(−1)=2

vmin=1−(cost)max=1−1=0

Hence, v lies between 0 and 2.

For acceleration 

a=dtdv=−sint

When t=0;x=0,v=0,a=0

When t=2π;x= positive , v=0,a

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Answer 2

Answer:

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