Question Statement

A particle is moving with constant acceleration $a$ . It starts with an initial velocity $v_{i}$ and after time $t$ , it covers a distance $S$ , reaching a final velocity $v_{f}$ . Prove that:

(i) $v_{f} = v_{i} + a t$

(ii) $S = v_{i} t + \frac{1}{2} a t^{2}$

(iii) $2 a S = v_{f}^{2} - v_{i}^{2}$

Background and Explanation

This problem derives the three standard kinematic equations for motion with constant acceleration (also known as the equations of motion). These equations relate the physical quantities of displacement, velocity, acceleration, and time, forming the foundation of classical mechanics for uniformly accelerated motion.

Solution

Part (i): First Equation of Motion

We begin with the definition of acceleration. Acceleration is defined as the rate of change of velocity with respect to time:

$a = \frac{Change in velocity}{Time taken} = \frac{v _{f} - v _{i}}{t}$

Rearranging this equation to solve for the final velocity $v_{f}$ :

$v_{f} - v_{i} = a t$

$v_{f} = v_{i} + a t$

This is the first equation of motion, showing that the final velocity equals the initial velocity plus the product of acceleration and time.

Part (ii): Second Equation of Motion

For an object moving with constant acceleration, the average velocity over a time interval is the arithmetic mean of the initial and final velocities:

$v_{a v} = \frac{v _{i} + v _{f}}{2}$

The total distance traveled equals the average velocity multiplied by the time interval:

$S = v_{a v} \cdot t = \frac{v _{i} + v _{f}}{2} \cdot t$

Now substitute the expression for $v_{f}$ from part (i) ( $v_{f} = v_{i} + a t$ ) into this equation:

$S = \frac{v _{i} + ( v _{i} + a t )}{2} \cdot t$

Simplify the numerator by combining like terms:

$S = \frac{2 v _{i} + a t}{2} \cdot t$

Distribute the multiplication by $t$ and simplify the fractions:

$S = \frac{2 v _{i}}{2} \cdot t + \frac{a t}{2} \cdot t$

$S = v_{i} t + \frac{1}{2} a t^{2}$

This is the second equation of motion, which allows us to calculate displacement without needing to know the final velocity explicitly.

Part (iii): Third Equation of Motion

We again start with the relationship between distance, average velocity, and time:

$S = v_{a v} \cdot t = \frac{v _{i} + v _{f}}{2} \cdot t$

From the first equation of motion (part i), we can express the time $t$ in terms of the velocities and acceleration:

$t = \frac{v _{f} - v _{i}}{a}$

Substitute this expression for $t$ into the distance equation:

$S = \frac{v _{i} + v _{f}}{2} \cdot \frac{v _{f} - v _{i}}{a}$

Multiply the numerators and denominators:

$S = \frac{( v _{f} + v _{i} ) ( v _{f} - v _{i} )}{2 a}$

Apply the algebraic identity $(a + b) (a - b) = a^{2} - b^{2}$ to the numerator:

$S = \frac{v _{f}^{2} - v _{i}^{2}}{2 a}$

Finally, multiply both sides by $2 a$ to obtain the standard form:

$2 a S = v_{f}^{2} - v_{i}^{2}$

This is the third equation of motion, which relates the velocities and displacement without involving the time variable explicitly.

Key Formulas or Methods Used

Definition of acceleration: $a = \frac{Δ v}{Δ t} = \frac{v _{f} - v _{i}}{t}$
Average velocity for constant acceleration: $v_{a v} = \frac{v _{i} + v _{f}}{2}$
Distance as product of average velocity and time: $S = v_{a v} \cdot t$
Difference of squares identity: $(a + b) (a - b) = a^{2} - b^{2}$

Summary of Steps

First equation: Start with the definition of acceleration $a = \frac{v _{f} - v _{i}}{t}$ and rearrange algebraically to isolate $v_{f}$
Second equation: Calculate distance using $S = v_{a v} t = \frac{v _{i} + v _{f}}{2} t$ , substitute the expression for $v_{f}$ from step 1, and simplify the resulting algebraic expression
Third equation: Express time as $t = \frac{v _{f} - v _{i}}{a}$ from step 1, substitute into $S = \frac{v _{i} + v _{f}}{2} t$ , then apply the difference of squares formula and rearrange to eliminate the fraction

Question Statement

A particle is moving with constant acceleration $a$ . It starts with an initial velocity $v_{i}$ and after time $t$ , it covers a distance $S$ , reaching a final velocity $v_{f}$ . Prove that:

(i) $v_{f} = v_{i} + a t$

(ii) $S = v_{i} t + \frac{1}{2} a t^{2}$

(iii) $2 a S = v_{f}^{2} - v_{i}^{2}$

Background and Explanation

Solution

Part (i): First Equation of Motion

We begin with the definition of acceleration. Acceleration is defined as the rate of change of velocity with respect to time:

$a = \frac{Change in velocity}{Time taken} = \frac{v _{f} - v _{i}}{t}$

Rearranging this equation to solve for the final velocity $v_{f}$ :

$v_{f} - v_{i} = a t$

$v_{f} = v_{i} + a t$

This is the first equation of motion, showing that the final velocity equals the initial velocity plus the product of acceleration and time.

Part (ii): Second Equation of Motion

For an object moving with constant acceleration, the average velocity over a time interval is the arithmetic mean of the initial and final velocities:

$v_{a v} = \frac{v _{i} + v _{f}}{2}$

The total distance traveled equals the average velocity multiplied by the time interval:

$S = v_{a v} \cdot t = \frac{v _{i} + v _{f}}{2} \cdot t$

Now substitute the expression for $v_{f}$ from part (i) ( $v_{f} = v_{i} + a t$ ) into this equation:

$S = \frac{v _{i} + ( v _{i} + a t )}{2} \cdot t$

Simplify the numerator by combining like terms:

$S = \frac{2 v _{i} + a t}{2} \cdot t$

Distribute the multiplication by $t$ and simplify the fractions:

$S = \frac{2 v _{i}}{2} \cdot t + \frac{a t}{2} \cdot t$

$S = v_{i} t + \frac{1}{2} a t^{2}$

This is the second equation of motion, which allows us to calculate displacement without needing to know the final velocity explicitly.

Part (iii): Third Equation of Motion

We again start with the relationship between distance, average velocity, and time:

$S = v_{a v} \cdot t = \frac{v _{i} + v _{f}}{2} \cdot t$

From the first equation of motion (part i), we can express the time $t$ in terms of the velocities and acceleration:

$t = \frac{v _{f} - v _{i}}{a}$

Substitute this expression for $t$ into the distance equation:

$S = \frac{v _{i} + v _{f}}{2} \cdot \frac{v _{f} - v _{i}}{a}$

Multiply the numerators and denominators:

$S = \frac{( v _{f} + v _{i} ) ( v _{f} - v _{i} )}{2 a}$

Apply the algebraic identity $(a + b) (a - b) = a^{2} - b^{2}$ to the numerator:

$S = \frac{v _{f}^{2} - v _{i}^{2}}{2 a}$

Finally, multiply both sides by $2 a$ to obtain the standard form:

$2 a S = v_{f}^{2} - v_{i}^{2}$

This is the third equation of motion, which relates the velocities and displacement without involving the time variable explicitly.

Key Formulas or Methods Used

Definition of acceleration: $a = \frac{Δ v}{Δ t} = \frac{v _{f} - v _{i}}{t}$
Average velocity for constant acceleration: $v_{a v} = \frac{v _{i} + v _{f}}{2}$
Distance as product of average velocity and time: $S = v_{a v} \cdot t$
Difference of squares identity: $(a + b) (a - b) = a^{2} - b^{2}$

Summary of Steps

First equation: Start with the definition of acceleration $a = \frac{v _{f} - v _{i}}{t}$ and rearrange algebraically to isolate $v_{f}$
Second equation: Calculate distance using $S = v_{a v} t = \frac{v _{i} + v _{f}}{2} t$ , substitute the expression for $v_{f}$ from step 1, and simplify the resulting algebraic expression
Third equation: Express time as $t = \frac{v _{f} - v _{i}}{a}$ from step 1, substitute into $S = \frac{v _{i} + v _{f}}{2} t$ , then apply the difference of squares formula and rearrange to eliminate the fraction

5.2 Q-13

Question Statement

Background and Explanation

Solution

Part (i): First Equation of Motion

Part (ii): Second Equation of Motion

Part (iii): Third Equation of Motion

Key Formulas or Methods Used

Summary of Steps

5.2 Q-13

Question Statement

Background and Explanation

Solution

Part (i): First Equation of Motion

Part (ii): Second Equation of Motion

Part (iii): Third Equation of Motion

Key Formulas or Methods Used

Summary of Steps