Ncert Exercises & Solutions

If in one-dimensional motion, instantaneous speed $v$ satisfies $0\leq v<v_0$, then:

1.	the displacement in time $T$ must always take non-negative values.
2.	the displacement $x$ in time $T$ satisfies $-\mathrm{v_0T} \lt x \lt \mathrm{v_0T}$.
3.	the acceleration is always a non-negative number.
4.	the motion has no turning points.

Hint: The velocity and displacement are vector quantities.

Step 1: Find the maximum value of the velocity in two directions.

For maximum and minimum displacement, we have to keep in mind the magnitude and direction of maximum velocity.

As maximum velocity in the positive direction is $v_{0}$ maximum velocity in opposite direction is also $v_{0}$ .

Step 2: Find the maximum and minimum displacement.

Maximum displacement in one direction $= v_{0} T$

Maximum displacement in opposite directions $= - v_{0} T$

Hence, $- v_{0} T < x < v_{0} T$

Hence, option (2) is the correct answer.

1.	the displacement in time \(T\) must always take non-negative values.
2.	the displacement \(x\) in time \(T\) satisfies \(-\mathrm{v_0T} \lt x \lt \mathrm{v_0T}\).
3.	the acceleration is always a non-negative number.
4.	the motion has no turning points.