MCQ
For the one-dimensional motion, described by $\text{x}=\text{t}-\sin\text{t}.$
- Ax (t) > 0 for all t > 0.
- Bv (t) > 0 for all t > 0.
- Ca (t) > 0 for all t > 0.
- Dv (t) lies between 0 and 2.
Explanation:
Position of the particle is given as a function of time i.e. $\text{x}=\text{t}-\sin\text{t}$ By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.
Velocity $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[\text{t}-\sin\text{t}]=1-\cos\text{t}$
If we again differentiate this equation w,r,t, time we will get will get acceleration of the particle as a function of time.
Acceleration $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[1-\cos\text{t}]=\sin\text{t}$
As acceleration a > 0 for all t > 0
Hence, x(t) > 0 for all t > 0
Velocity $\text{v}=1-\cos\text{t}$
When, $\cos\text{t}-1,$ velocity v = 0
$\text{v}_\text{max}=1-(\cos\text{t})_\text{min}=1-(-1)=2$
$\text{v}_\text{min}=1-(\cos\text{t})_\text{max}=1-1=0$
Hence, v lies between 0 and 2.
Acceleration $\text{a}=\frac{\text{dv}}{\text{dt}}=-\sin\text{t}$
When t = 0; x = 0, v = 0, a = 0
When $\text{t}=\frac{\pi}{2};$ x = positive, v = 0, a = -1 (negative)
When $\text{t}=2\pi,\text{x}=0,\text{v}=0,\text{a}=0$
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