Class 11 NCERT PHYSICS • Chapter 3: Motion in a Plane • Question 20

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We need to determine which of the provided kinematic relations remain valid for any arbitrary motion in space.

Concept: Average velocity

The definitions of average velocity and average acceleration are fundamental and hold for any type of motion. In contrast, equations derived for constant acceleration (uniform motion) are only valid when the acceleration vector does not change with time.

Given:

$v_{\text{average}} = \frac{1}{2}\,(v(t_1) + v(t_2))$
$v_{\text{average}} = \frac{r(t_2) - r(t_1)}{t_2 - t_1}$
$v(t) = v(0) + at$
$r(t) = r(0) + v(0)t + \frac{1}{2}at^2$
$a_{\text{average}} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$

Solving:

Step 1: Evaluate relations (1), (3), and (4).
These relations are specifically derived for motion with constant acceleration. For arbitrary motion, the acceleration $a$ may be a function of time, $a(t)$ . In such cases, the velocity does not change linearly, and the position does not follow a simple quadratic form. Therefore, these relations are not generally true.
Step 2: Evaluate relation (2).
By definition, the average velocity over a time interval is the total displacement divided by the total time taken:

v_{\text{average}} = \frac{\Delta r}{\Delta t} = \frac{r(t_2) - r(t_1)}{t_2 - t_1}

This is a universal definition and is true for any arbitrary motion.

Step 3: Evaluate relation (5).
The expression

\frac{v(t_2) - v(t_1)}{t_2 - t_1}

represents the change in velocity over the time interval, which is the definition of average acceleration:

a_{\text{average}} = \frac{\Delta v}{\Delta t} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}

Like average velocity, this definition is universally true for all types of motion.

Answer

2 \text{ and } 5

NCERT Solutions

Motion in a Plane

For any arbitrary motion in space, which of the following relations are true:

1. $v_ \text{average}=\frac{1}{2}\,(v(t_1 )+v(t_2 ))$

2. $v_ \text{average}=\frac{[r(t_2 )−r(t_1 )]}{(t_2 − t_1 )}$

3. $v(t)=v(0)+at$

4. $r(t)=r(0)+v(0)t+(1/2)at^2$

5. $a_ \text{average}=[v(t_2 )−v(t_1 )]/(t_2 − t_1 )$

(The ‘average’ stands for average of the quantity over the time interval $t_1$ to $t_2$ )

SOLUTION

We need to determine which of the provided kinematic relations remain valid for any arbitrary motion in space.

Concept: Average velocity

Given:

Solving:

Answer

Motion in a Plane

For any arbitrary motion in space, which of the following relations are true:

1. vaverage=12 (v(t1)+v(t2))v_ \text{average}=\frac{1}{2}\,(v(t_1 )+v(t_2 ))vaverage​=21​(v(t1​)+v(t2​))

2. vaverage=[r(t2)−r(t1)](t2−t1)v_ \text{average}=\frac{[r(t_2 )−r(t_1 )]}{(t_2 − t_1 )}vaverage​=(t2​−t1​)[r(t2​)−r(t1​)]​

3. v(t)=v(0)+atv(t)=v(0)+atv(t)=v(0)+at

4. r(t)=r(0)+v(0)t+(1/2)at2 r(t)=r(0)+v(0)t+(1/2)at^2r(t)=r(0)+v(0)t+(1/2)at2

5. aaverage=[v(t2)−v(t1)]/(t2−t1)a_ \text{average}=[v(t_2 )−v(t_1 )]/(t_2 − t_1 )aaverage​=[v(t2​)−v(t1​)]/(t2​−t1​)

(The ‘average’ stands for average of the quantity over the time interval t1t_1t1​ to t2t_2t2​)

SOLUTION

We need to determine which of the provided kinematic relations remain valid for any arbitrary motion in space.

Concept: Average velocity

Given:

Solving:

Answer

1. $v_ \text{average}=\frac{1}{2}\,(v(t_1 )+v(t_2 ))$

2. $v_ \text{average}=\frac{[r(t_2 )−r(t_1 )]}{(t_2 − t_1 )}$

3. $v(t)=v(0)+at$

4. $r(t)=r(0)+v(0)t+(1/2)at^2$

5. $a_ \text{average}=[v(t_2 )−v(t_1 )]/(t_2 − t_1 )$

(The ‘average’ stands for average of the quantity over the time interval $t_1$ to $t_2$ )