Equations of Motion

Kinematics Equations Of Motion

The equations we’re discussing are fundamental to understanding motion, specifically when dealing with constant or uniform acceleration. Known as the equations of motion, they form the cornerstone of kinematics, a branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies without considering the forces that cause them to move.

They are only valid if the acceleration is CONSTANT (UNIFORM acceleration).

Equation 1 – $v = u + at$

$$v = u + at \tag{1}$$

This equation calculates the final velocity ($v$) of an object when you know its initial velocity ($u$), the acceleration ($a$), and the time ($t$) over which the acceleration is applied.

Equation 2 – $s = \frac{1}{2} (u + v) t$

$$s = \frac{1}{2} (u + v) t \tag{2}$$

This allows you to calculate displacement using the average of the initial and final velocities over a given time period.

Equation 3 – $s = ut +\frac{1}{2}at^{2}$

$$s = ut + \frac{1}{2} at^{2} \tag{3}$$

This allows you to determine the displacement ($s$) of an object, factoring in its initial velocity, the acceleration, and the time period of motion.

Equation 4 – $v^{2} = u^{2} + 2as$

$$v^{2} = u^{2} + 2as \tag{4}$$

This equation relates the square of the final velocity to the square of the initial velocity, the acceleration, and the displacement.


$$v = \text{final velocity}$$

$$u = \text{ initial velocity}$$

$$a = \text{ acceleration}$$

$$t = \text{ change in time}$$

$$s = \text{ change in displacement}$$

NOTE: These equations are ONLY valid IF AND ONLY IF acceleration is constant. (Very Important!)

How To Select Which Equation Of Motion To Use?

You select the equations based on the amount of information that you have.

Information that you haveEquations of motion to use
$v = u + at$
$s = ut + \frac{1}{2} at^{2}$
$s = \frac{1}{2} (u + v)$
$v^{2} = u^{2} + 2as$

If you prefer to read words, the information in the table can be found (in words) below.

$$v = u + at$$

The above equation is suitable to use if you do not have the displacement.

$$s = ut + \frac{1}{2} at^{2}$$

The above equation is suitable to use if you do not have the final velocity.

$$s = \frac{1}{2} (u + v) t$$

The above equation is suitable to use if you do not have the acceleration.

$$v^{2} = u^{2} + 2as$$

The above equation is suitable to use if you do not have the time taken.

[A Level] Derivation Of Equations Of Motion

Equation 1 – $v = u + at$

Remember that equation for acceleration is the following:

$$a = \frac{v -u}{t}$$

You can re-arrange the equation to give:

$$v = u + at$$

Equation 2 – $s = \frac{1}{2} (u + v) t$

The velocity of a body moving with uniform acceleration increases steadily. Its average velocity therefore equals half the sum of its initial and final velocities:

$$\text{Average velocity} = \frac{u+v}{2}$$

If $s$ is the distance moved in time $t$, then since $\text{average velocity} = \frac{\text{distance}}{\text{time}} = \frac{s}{t}$,

$$\frac{s}{t} = \frac{u+v}{2}$$

which re-arranges to give:

$$s = \frac{1}{2} (u + v) t$$

Equation 3 – $s = ut +\frac{1}{2}at^{2}$

Substitute the first equation into the second equation.

You will get the following:

$$s = ut +\frac{1}{2}at^{2}$$

Equation 4 – $v^{2} = u^{2} + 2as$

Re-arranging the first equation to the following:

$$t = \frac{v-u}{a}$$

Substitute the above equation into the third equation and you will get:

$$v^{2} = u^{2} + 2as$$

Worked Examples

Example 1

A boy throws a ball vertically upwards. The ball falls back into his hand 2.0 seconds later. Neglecting air resistance, what is the speed at which the ball leaves his hand?

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Taking upward motion as positive,

$$ \begin{aligned} s &= ut + \frac{1}{2} at^2 \\ 0 &= u(2) + \frac{1}{2} (-9.81)(2^2) \\ u &= \frac{1}{2} (9.81)(2) \\ u &= 9.81 \, \text{m/s}^{-1} \end{aligned}$$

Example 2: The Racing Car

Question: A racing car accelerates from rest at a uniform acceleration of $4 \, \text{m/s}^2$ for $3$ seconds. How far does the car travel in this time?

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  • Initial velocity ($u$) = $0 \, \text{m/s}$ (since the car starts from rest)
  • Acceleration ($a$) = $4 \, \text{m/s}^2$
  • Time ($t$) = $3 \, \text{s}$

We’ll use the equation $s = ut + \frac{1}{2} at^2$:

$s = 0 \times 3 + \frac{1}{2} \times 4 \times 3^2 = 0 + 18 = 18 \, \text{m}$

The car travels 18 meters.

Example 3: The Leap of Faith

Question: A stuntman jumps off a moving motorcycle to land on a moving train. The motorcycle is traveling at $20 \, \text{m/s}$, and the train is accelerating away at $2 \, \text{m/s}^2$. If the motorcycle and the train are $30 \, \text{m}$ apart when the stuntman jumps, how long does he have before the train is too far away? Assume the stuntman lands with the same speed as his jump.

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  • Initial relative velocity of the stuntman to the train ($u$) = $20 \, \text{m/s}$ (assuming the train is initially at this speed or less for the stuntman to catch up)
  • Relative acceleration ($a$) = $-2 \, \text{m/s}^2$ (since the train accelerates away from the stuntman, reducing his relative speed towards the train)
  • Initial separation ($s$) = $30 \, \text{m}$

We can solve for the time ($t$) it takes for this distance to close, which requires using the equation:

$$\begin{aligned} s &= ut + \frac{1}{2} at^2 \\ 30 &= 20t + \frac{1}{2} \times (-2) \times t^{2} \\ 0 &= -t^2 +20t-30 \\ t^{2}-20t+30 &=0 \end{aligned}$$

Solving this quadratic equation for $t$:

$$t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

where $a = 1$, $b = -20$, and $c = 30$.

The quadratic equation yields two solutions: $t \approx 18.37$ seconds and $t \approx 1.63$ seconds. The smaller value, $1.63$ seconds, is the relevant solution here, as it represents the time it would take for the stuntman to potentially catch up to the train before the increasing separation due to the train’s acceleration makes it too far to reach.

Therefore, the stuntman has approximately 1.63 seconds to land on the train before it becomes too far away.

Example 4: The Dropped Object from a Tower

Question: An object is dropped from the top of a $45 \, \text{m}$ high tower. How long does it take to hit the ground? (Ignore air resistance, $g = 9.8 \, \text{m/s}^2)$

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  • Initial velocity ($u$) = $0$ (since the object is dropped)
  • Acceleration ($a$) = $9.8 \, \text{m/s}^2$ (acceleration due to gravity)
  • Distance ($s$) = $45 \, \text{m}$

Using $s = ut + \frac{1}{2} at^2$:

$45 = 0 + \frac{1}{2} \times 9.8 \times t^2$

Solving for $t$, $t^2 = \frac{45 \times 2}{9.8}$, $t \approx 3.03 \, \text{s}$.

It takes approximately 3.03 seconds for the object to hit the ground.

Example 5: The Speeding Bullet

Question: A bullet is fired straight up into the air with an initial velocity of $490 \, \text{m/s}$. How high does the bullet go before stopping momentarily? (Use $g = 9.8 \, \text{m/s}^2$ for gravity)

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  • Initial velocity ($u$) = $490 \, \text{m/s}$
  • Final velocity ($v$) = $0 \, \text{m/s}$ (at the highest point)
  • Acceleration ($a$) = $-9.8 \, \text{m/s}^2$ (since gravity is acting against the motion)

Using $v^2 = u^2 + 2as$:

$$0 = 490^2 + 2 \times (-9.8) \times s$$

Solving for $s$, $s = \frac{490^2}{2 \times 9.8}$, $s \approx 12250 \, \text{m}$.

The bullet goes up to approximately 12,250 meters before stopping momentarily.

Example 6: The Horizontal Projectile

Question: A ball is thrown horizontally from a cliff $80 \, \text{m}$ high with an initial speed of $15 \, \text{m/s}$. How far from the base of the cliff will the ball land? (Ignore air resistance, $g = 9.8 \, \text{m/s}^2)$

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  • Height of the cliff ($s$) = $80 \, \text{m}$
  • Initial horizontal velocity ($u$) = $15 \, \text{m/s}$
  • Acceleration due to gravity ($a$) = $9.8 \, \text{m/s}^2$

First, find the time it takes to hit the ground using $s = \frac{1}{2} gt^2$:

$$80 = \frac{1}{2} \times 9.8 \times t^2$$

Solving for $t$, $t^2 = \frac{80 \times 2}{9.8}$, $t \approx 4.04 \, \text{s}$.

Then, calculate the horizontal distance using $\text{distance} = ut$:

$$\text{distance} = 15 \times 4.04 \approx 60.6 \, \text{m}$$

The ball will land approximately 60.6 meters from the base of the cliff.

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