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Instantaneous Velocity

Statement-1: ...

Question

Statement-1: If displacement is a linear function of time, its average and instantaneous velocity will be same.
Statement-2: If the acceleration of a moving particle is zero, the particle continues to move along the same direction.

A

Both statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1

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B

Both statement 1 and statement 2 are correct and statement 2 is not the correct explanation of statement 1.

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C

Statement 1 is correct and statement 2 is incorrect.

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D

Statement 1 is incorrect and statement 2 is correct.

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Solution

The correct option is B Both statement 1 and statement 2 are correct and statement 2 is not the correct explanation of statement 1.
If displacement is a linear function of time then velocity is constant i.e. motion will be of uniform nature. The slope of the graph will always be same at every point, so instantaneous velocity and average velocity will be equal.
If acceleration is $0$ , then change is velocity is $0$ which means velocity is constant throughout and the particle continues to move along the same direction.

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Similar questions

Q. Pick the correct statements:
(a) Average speed of a particle in a given time is never less than the magnitude of the average velocity.
(b) It is possible to have a situation in which

|\frac{d \vec{v}}{d t}| \neq 0 but \frac{d}{d t} = |\vec{v}| = 0

(c) The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero is the interval.
(d) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval. (Infinite accelerations are not allowed).

Q. Statement-1: If displacement is a linear function of time, its average and instantaneous velocity will be same.
Statement-2: If the acceleration of a moving particle is zero, the particle continues to move along the same direction.

If a particle is moving in such a way that it's average acceleration turns out to be different for a number of different time intervals, the particle is said to have variable acceleration. The acceleration can vary in magnitude, or in direction or both. In such cases we find acceleration at any instant, called the instantaneous acceleration. It is defined as

\to a = lim Δ t = 0 \frac{Δ \to v}{Δ t} = \frac{d \to v}{d t}

That is acceleration of a particle at time

t

is the limiting value of

\frac{Δ v}{Δ t}

t

Δ t

approaches zero. The direction of the instantaneous acceleration

\to a

is the limiting direction of the vector in velocity

Δ v

.

A particle is moving along a straight line with

10 m s^{- 1}

. It takes a U-turn in

5 s

and continues to move along with the same velocity

10 m s^{- 1}

. Find the magnitude of average acceleration during turning.

Statement I For a particle moving along a straight line, the average velocity vector over a time interval would be equal to instantaneous velocity vector at the end of the interval only if the particle is moving with constant velocity.

Statement II If a particle is moving with constant velocity, then average velocity for any time interval is equal to the velocity of particle.

Q. A particle moving along a straight line with constant acceleration is having initial and final velocity as

5 m / s

15 m / s

respectively in a time interval of

5 s

. Find the distance travelled by the particle and the acceleration of the particle. If the particle continues with same acceleration, find the distance covered by the particle in the

8 t h

second of its motion (direction of motion remains same).

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