For the one-dimensional motion, describe by x = t – sint a) x(t)>0 for all t>0 b) v(t)>0 for all t>0 c) a(t)>0 for all t>0 d) v(t) lies between 0 and 2
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=−1 (negative) When t=π,x= positive, v= positive , a=0 When t=2π,x=0,v=0,a=0
