Kinematic Equations of Motion - Ninth Grade Science
Kinematics reaches its mathematical pinnacle when we synthesize position, time, velocity, and acceleration into a unified set of predictive algebraic relationships known as the kinematic equations of motion. These three foundational formulas allow you to predict the future position and velocity of any moving object, or reconstruct its past motion, provided the object moves with constant, uniform acceleration. Whether designing runway lengths for commercial jetliners, calculating the stopping distances of motor vehicles, or modeling the trajectory of a rocket, the kinematic equations are indispensable tools in physical science. In this chapter, you will derive each equation step by step, analyze their physical meanings, and master problem-solving strategies.
The Standard Variables of Kinematics
To work fluently with the equations of motion, you must become intimately familiar with the five standard kinematic variables, often remembered collectively by the acronym SUVAT.
+----------+-----------------------------+------------------------+
| Variable | Physical Meaning | Standard SI Unit |
+----------+-----------------------------+------------------------+
| s | Displacement (or distance) | Meter (m) |
| u | Initial velocity | Meter per second (m/s) |
| v | Final velocity | Meter per second (m/s) |
| a | Uniform acceleration | Meters/sec^2 (m/s^2) |
| t | Time interval elapsed | Second (s) |
+----------+-----------------------------+------------------------+
Each equation connects four of these five variables. When solving any mechanics problem, your first step is always to identify the three known quantities, determine the single unknown quantity you need to find, and select the unique equation linking those four parameters.
Derivation of the First Equation of Motion: $v = u + at$
The first equation of motion relates velocity, acceleration, and time. It provides a direct formula to compute the final velocity of an accelerating body after a known time interval.
Derivation from Definition
Acceleration = (Change in Velocity) / Time
a = (v - u) / t
Multiply both sides by t:
a * t = v - u
Rearrange to isolate v:
v = u + at
Physical Interpretation
The term $u$ represents the baseline velocity the object possessed before acceleration began. The term $at$ represents the additional velocity accumulated (or lost) due to acceleration acting continuously over duration $t$. Thus, the equation states that the final velocity equals the original velocity plus the cumulative velocity change. If an object starts from rest, $u = 0$, and the equation reduces to $v = at$.
Derivation of the Second Equation of Motion: $s = ut + \frac{1}{2}at^2$
The second equation of motion relates displacement, initial velocity, uniform acceleration, and time. It calculates how far an object travels during an accelerating interval.
1. Average Velocity for Uniform Acceleration:
v_avg = (u + v) / 2
2. Displacement = Average Velocity * Time:
s = [ (u + v) / 2 ] * t
3. Substitute First Equation (v = u + at) into Displacement formula:
s = [ (u + (u + at)) / 2 ] * t
s = [ (2u + at) / 2 ] * t
s = [ u + (1/2)at ] * t
4. Expand the expression:
s = ut + (1/2)at^2
Physical Interpretation
The displacement consists of two distinct components. The first term, $ut$, is the displacement the object would have covered if it had simply continued moving at its initial velocity $u$ without any acceleration. The second term, $\frac{1}{2}at^2$, represents the extra displacement gained (or lost) as a direct consequence of the acceleration. Notice that this additional displacement scales with the square of time: doubling the accelerating time quadruples this component of displacement.
Derivation of the Third Equation of Motion: $v^2 = u^2 + 2as$
The third equation of motion is often called the timeless equation because it links initial velocity, final velocity, acceleration, and displacement without requiring knowledge of the elapsed time $t$.
1. From First Equation, isolate time t:
t = (v - u) / a
2. Recall Displacement from Average Velocity:
s = [ (v + u) / 2 ] * t
3. Substitute t = (v - u) / a into the Displacement formula:
s = [ (v + u) / 2 ] * [ (v - u) / a ]
s = (v^2 - u^2) / (2a)
4. Multiply both sides by 2a:
2as = v^2 - u^2
5. Rearrange to isolate v^2:
v^2 = u^2 + 2as
Physical Interpretation
This equation directly foreshadows the work-energy theorem in dynamics. If you multiply both sides of $v^2 - u^2 = 2as$ by half the mass ($\frac{1}{2}m$), you obtain $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mas = Fs$, which proves that the change in kinetic energy equals the work done by the net force.
Summary of the Three Kinematic Equations
+-----------------------+-----------------------------+---------------------+
| Equation Number | Mathematical Formula | Missing Variable |
+-----------------------+-----------------------------+---------------------+
| First Equation | v = u + at | Displacement (s) |
| Second Equation | s = ut + (1/2)at^2 | Final Velocity (v) |
| Third Equation | v^2 = u^2 + 2as | Time (t) |
+-----------------------+-----------------------------+---------------------+
Crucial Limitation: Constant Acceleration Only
These three equations are valid if and only if acceleration $a$ is strictly constant throughout the entire interval of motion. If acceleration varies with time or position—such as a rocket whose mass decreases dramatically as fuel burns or a car navigating stop-and-go traffic—these algebraic equations fail, and integral calculus must be used instead.
Motion Under Gravity: Special Case of Kinematic Equations
When an object moves vertically under the exclusive influence of Earth's gravity, we replace horizontal displacement $s$ with vertical height $h$, and replace linear acceleration $a$ with gravitational acceleration $g$ (or $-g$ depending on coordinate orientation).
Upward Motion (Taking Up as Positive, a = -g):
1. v = u - gt
2. h = ut - (1/2)gt^2
3. v^2 = u^2 - 2gh
Downward Free Fall from Rest (Taking Down as Positive, u = 0, a = +g):
1. v = gt
2. h = (1/2)gt^2
3. v^2 = 2gh ===> v = sqrt(2gh)
Chapter Exercises
Exercise 1: An airplane starting from rest accelerates uniformly down a runway at 3.0 meters per second squared. If the required takeoff speed for the aircraft is 75 meters per second, calculate the minimum runway length needed for the plane to safely become airborne, and determine the time elapsed from rest to takeoff.
Exercise 2: A car traveling at 90 kilometers per hour detects an obstacle on the roadway. The driver reacts, and the brakes apply a uniform deceleration of 5.0 meters per second squared. Calculate the total stopping distance of the car from the moment the brakes engage.
Exercise 3: A stone is dropped from the top of an observation tower and strikes the ground 4.0 seconds later. Assuming air resistance is negligible and taking the acceleration due to gravity as 9.8 meters per second squared, calculate the height of the tower and determine the velocity of the stone immediately before striking the ground.
Solutions and Explanations
Solution 1: Identify the given kinematic variables: initial velocity $u = 0$ m/s, final velocity $v = 75$ m/s, and acceleration $a = 3.0 ext{ m/s}^2$. To find the minimum runway length (displacement $s$), use the third kinematic equation since time is not needed: $v^2 = u^2 + 2as$. Substituting the knowns gives $(75)^2 = (0)^2 + 2(3.0)s$, which simplifies to $5625 = 6.0s$. Solving for $s$ yields $s = 5625 / 6.0 = 937.5$ meters. To find the time elapsed, use the first kinematic equation: $v = u + at$, which gives $75 = 0 + 3.0t$. Solving for $t$ yields $t = 75 / 3.0 = 25$ seconds.
Solution 2: First convert the vehicle's initial speed from kilometers per hour to meters per second: 90 multiplied by 5 divided by 18 equals 25 meters per second. The initial velocity $u$ is 25 m/s, the final velocity $v$ when stopped is 0 m/s, and the acceleration $a$ is negative 5.0 meters per second squared due to deceleration. We want to find the stopping distance $s$. Using the third kinematic equation, $v^2 = u^2 + 2as$, substitute the values: $(0)^2 = (25)^2 + 2(-5.0)s$. This yields $0 = 625 - 10.0s$. Rearranging gives $10.0s = 625$, which results in $s = 62.5$ meters. The car requires 62.5 meters to come to a complete stop.
Solution 3: The stone is dropped from rest, so initial velocity $u = 0$ m/s, time $t = 4.0$ s, and downward acceleration $g = 9.8 ext{ m/s}^2$. To find the height of the tower $h$, use the second kinematic equation adapted for free fall: $h = ut + \frac{1}{2}gt^2$. Substituting the values gives $h = (0)(4.0) + \frac{1}{2}(9.8)(4.0)^2 = 0 + 4.9 imes 16 = 78.4$ meters. To determine the velocity immediately before impact, use the first kinematic equation: $v = u + gt = 0 + 9.8 imes 4.0 = 39.2$ meters per second downward.