In one dimensional motion, instantaneous speed v satisfies 0 <v_0… - Vidyadip

MCQ

In one dimensional motion, instantaneous speed $v$ satisfies $0\le\text{v}<\text{v}_0.$

A
The displacement in time $T$ must always take non$-$negative values.
✓
The displacement $x$ in time $T$ satisfies $\text{v}_0\text{ T}<\text{x}<\text{v}_0\text{ T}.$
C
The acceleration is always a non$-$negative number.
D
The motion has no turning points.

✓

Answer

Correct option: B.

The displacement $x$ in time $T$ satisfies $\text{v}_0\text{ T}<\text{x}<\text{v}_0\text{ T}.$

Key concept: Instantaneous speed: It is the speed of a particle at a particular instant of time.
When we say “speed”, it usually means instantaneous speed.
The instantaneous speed is average speed for infinitesimally small time interval $($i.e., $\Delta>0).$
Thus, Instantaneous speed $\text{v}=\DeclareMathOperator*{\median}{\text{lim}} \median\limits_{\Delta\text{t}\rightarrow0}\frac{\Delta\text{s}}{\Delta\text{t}}=\frac{\text{ds}}{\text{dt}}$
As instantaneous speed is less than maximum speed.
Then either the velocity is increasing or it is decreasing.
For maximum and minimum displacement we have to keep in mind the magnitude and direction of maximum velocity.
As maximum velocity in positive direction is $v_0$, magnitude of maximum velocity in opposite direction is also $v_0$.
Maximum displacement in one direction $=\text{v}_0\text{T}$ Maximum displacement in opposite directions $=-\text{v}_0\text{T}$
Hence, $-\text{v}_0\text{T}<\text{x}<\text{v}_0\text{T}.$

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