For the one-dimensional motion, described by x=t-sint . . .

Question : For the one-dimensional motion, described 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.

Doubt by Krish

Solution : 

a) x=t-sint

By putting different-different values of t in above equation we get corresponding values of x. 


When we plot the graph then we get something like this


From above, it is clear that for all values of t>0, we get the value of t-sint >0

Hence, the statement a) x(t)>0 for all t>0 is correct.

b)
x=t-sint
differentiating both sides w.r.t. time t.
v=dx/dt
v=d(t-sint)/dt
v=1-cost

By putting different-different values of t in above equation we get corresponding values of v. 
The corresponding graph will look like this



From above, it is clear that for all values of t>0, we get the value of 1-cost >0.

Hence, the statement a) v(t)>0 for all t>0 is correct.

c) We know, 
a=dv/dt
a=d(1-cost)/dt
a=0-(-sint)
a=sint

By putting different-different values of t in above equation we get corresponding values of a=sint. 
The corresponding graph will look like this


From the graph it is clear that for the some values of t there are -ve values of a(t).

Hence, the statement c) a(t)>0 for all t>0 is not correct.

d) v=1-cost

We know, the range of cosx varies between -1 and 1 so, 

-1≤cost≤1
Multiplying both sides by -1
1≥-cost≥-1
Adding both sides by 1
1+1≥1-cost≥1-1
2≥1-cost≥0
2≥v(t)≥0
0≤v(t)≤2

Hence, it is clear that the value of v(t) lies between 0 and 2 both included. 

Hence, d) v(t) lies between 0 and 2 is correct.