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2.1.4 Equations of Motion

By the end of this lesson you can:

  • Choose the correct equation of constant acceleration for a problem
  • Solve problems with an unknown displacement, time or velocity
  • Apply the equations to vertical motion under gravity
Find your course: NSW Year 11: Motion relationshipsNSW Year 12QLD Unit 2: Linear motion and forceVIC Unit 2: MotionAP Physics 1: Unit 1IB: A.1 Kinematics
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🎥Watch
Learn the idea: a short explanation, then a check.

In the previous lesson we looks at graphing motion.
If motion can be graphed and there are certain trends, then there are mathematical equation that describe that trend.

In this lesson we examine these mathematical equations.
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There are relationships between each of the variables involved in kinematics and these mathematical relationships are modelled in what are know as the kinematic equations, or equations of motion. 

Watch the video which covers the 5 forms of kinematic equations and how they are derived.

In the previous lesson, we looked at graphing motion and interpreting the relationships between displacement, velocity, and time.

When an object moves with constant (uniform) acceleration, we can model these graphical relationships using precise mathematical formulas known as the kinematic equations, or equations of motion.

Key Motion Variables (SUVAT):
  • s = Displacement (meters, m)
  • u = Initial velocity (meters per second, m/s)
  • v = Final velocity (meters per second, m/s)
  • a = Acceleration (meters per second squared, m/s2)
  • t = Time elapsed (seconds, s)

Depending on which variables are known and which variable you are solving for, you can select the appropriate equation below:

v = u + at
s = ut + ½at2
v2 = u2 + 2as
s = vt - ½at2
s = (u + v)⁄2 · t
Note: The two equations highlighted in red are less commonly used, but remain mathematically valid derivations.
How to Solve physics Problems
Before we look at a sample problem, there is a useful technique to use consistently when doing calculation type problems and it requires you to be RUDE.
​Read Understand Data Equation.
Watch the video as I explain
📝Practise
Worked examples first, then problems to try, then worksheets.
Sample Problem
We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.
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Check Your Understanding: Kinematics Practice

1. A car accelerates uniformly from rest (u = 0 m/s) at 3.0 m/s2 for 4.0 seconds. What is its final velocity?
2. Which kinematic equation should you select when the time variable (t) is neither given nor required?
3. A stone is dropped from rest off a cliff. Taking acceleration due to gravity as g = 9.8 m/s2, how far does it fall during the first 2.0 seconds?

Practice Problems: Kinematics & Motion

1. A car travels 20 km at 40 km/h. At what speed must the car travel for the next 20 km to achieve an average speed of 60 km/h for the entire 40 km journey?

Total Journey Target: Total distance = 40 km, Target average speed = 60 km/h

Total time needed = dtotal / vavg = 40 / 60 = 2/3 hour (40 minutes)

First Leg: t1 = 20 km / 40 km/h = 0.5 hour (30 minutes)

Second Leg: Remaining time t2 = 2/3 − 0.5 = 1/6 hour (10 minutes)

Required speed v2 = d2 / t2 = 20 km / (1/6 h) = 120 km/h

Answer: 120 km/h (120 km h-1)
2. A car leaves home at 8:00 AM traveling at 60 km/h. A second car leaves the same location at 8:30 AM traveling at 80 km/h in the same direction. When and where will the second car overtake the first car?

Let t = time in hours after Car 2 leaves (after 8:30 AM):

Car 1 head start (30 min = 0.5 h) distance = 60 km/h × 0.5 h = 30 km

Equating distances traveled from home:

30 + 60t = 80t

20t = 30 → t = 1.5 hours

Time: 8:30 AM + 1.5 hours = 10:00 AM

Distance: 80 km/h × 1.5 h = 120 km

Answer: 10:00 AM, 120 km from home
3. An aircraft needs to reach a take-off speed of 80 m/s. If it accelerates from rest at a constant 3.0 m/s2, calculate the minimum runway length required.

Given: u = 0 m/s, v = 80 m/s, a = 3.0 m/s2

Formula: v2 = u2 + 2as

Calculation:

(80)2 = 02 + 2(3.0)s

6400 = 6.0s

s = 6400 / 6.0 ≈ 1066.67 m

Answer: 1070 m (or 1.07 × 103 m)
4. An object moving with constant acceleration can certainly slow down. But can an object ever come to a permanent halt if its acceleration truly remains constant? Explain.

Explanation: No, an object cannot come to a permanent halt if its acceleration remains truly constant and non-zero.

If velocity reaches zero (v = 0), a non-zero constant acceleration a means the rate of change of velocity continues. Immediately after stopping, the object will start moving in the direction of the acceleration (reversing its initial direction of motion).

An object can only remain at a permanent halt if the net acceleration becomes zero once velocity reaches zero.

Answer: No. Continuous non-zero acceleration will cause the object to reverse direction immediately after stopping.
5. A car is moving at a uniform speed of 20 m/s when a cow appears on the road 60 m ahead. If the brakes apply a deceleration of 4.0 m/s2 and it takes 6.0 seconds total to bring the car to a rest (from the moment the cow is spotted), calculate:
(a) The driver's reaction time before applying the brakes.
(b) Does the car hit the cow?

(a) Reaction Time:

Braking time tbraking = (v − u) / a = (0 − 20) / (−4.0) = 5.0 seconds.

Total time = Reaction time (trxn) + Braking time (tbraking)

6.0 s = trxn + 5.0 s → trxn = 1.0 second.

(b) Total Stopping Distance:

Reaction distance drxn = 20 m/s × 1.0 s = 20 m.

Braking distance dbraking = ut + ½at2 = (20 × 5) + ½(−4)(5)2 = 100 − 50 = 50 m.

Total stopping distance = 20 m + 50 m = 70 m.

Since 70 m > 60 m, the car needs 10 m more than the gap to stop, so it cannot stop in time.

Answer: Reaction time = 1.0 second; Yes, the car hits the cow.
GOING FURTHER

Determining the equations of motion using graphical analysis or algebraic substitution are not the only ways to derive them. A far more fundamental and powerful approach is to use calculus.

In real-world motion, acceleration is rarely strictly constant. Calculus allows us to analyze non-uniform motion by connecting displacement x(t), velocity v(t), and acceleration a(t) through rates of change and accumulation.

1. Differentiation (Finding Rates of Change)

Taking the derivative with respect to time (t):

v(t) = dx⁄dt
a(t) = dv⁄dt = d2x⁄dt2

2. Integration (Accumulating Change)

Integrating over time yields state variables plus initial conditions:

v(t) = ∫ a(t) dt + v0
x(t) = ∫ v(t) dt + x0

In the video below, using both differential and integral calculus, I show step-by-step how the standard kinematic equations are derived from first principles.

Check Your Understanding: Calculus of Motion Practice

1. If an object's position as a function of time is given by x(t) = 3t2 − 4t + 2, what is its velocity at t = 3 seconds?
2. Integrating a constant acceleration a with respect to time t yields which kinematic relation?
3. An object starts from rest at t = 0 and experiences an acceleration a(t) = 6t m/s2. What is its velocity at t = 2 seconds?