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Kinematics

Kinematics focuses on the movement of objects. It deals with forces and the geometric aspects of motion, which is related to velocity and acceleration.

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# Kinematics

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Kinematics focuses on the movement of objects. It deals with forces and the geometric aspects of motion, which is related to velocity and acceleration.

Let's define some specific terms to understand the topic of kinematics:

• Distance is the total length covered irrespective of the direction of a motion.

• Displacement is the distance moved in a certain direction.

• Speed is the distance covered per unit of time.

• Velocity is the rate of change of displacement.

• Acceleration is the rate of change of velocity.

We can now use these to describe the motion of an object and the geometry of the curve it creates as it moves.

## Variable acceleration

In Variable Acceleration, acceleration changes over time. We need to use Differentiation and Integration to convert between displacement, velocity, and acceleration. We use Differentiation to convert from displacement to velocity and from velocity to acceleration. Then we use Integration to go back from acceleration to velocity and from velocity to displacement.

#### Example:

Find the times at which a particle is at instantaneous rest if its displacement is given as

Use differentiation to convert from displacement into velocity.

This is the velocity function. We want to know when v = 0

(3t -2) (t -4) = 0

The particle is at instantaneous at t = 4s or t = 2/3 s.

## Calculus kinematics

Calculus makes deriving two of the three Equations of motion easy. Since we know acceleration is the first derivative of velocity with respect to time, use the definition to reverse it. Instead of finding velocity, integrate it to find acceleration.

Various Functions describe the quantities of acceleration, displacement, and velocity. They can also be transformed into Functions that describe the other two quantities. There are two ways of doing this: differentiation (finding the derivative) or integration (finding the integral).

• Displacement s = f (t) [location in reference to an origin]

• Velocity [derivative of displacement]

• Acceleration [derivative of velocity]

Let's analyze these concepts together. You can understand the characteristics of kinematics based on the output of our function. For displacement, a value of 0 means our position is at the origin. If displacement is positive, then it's to the right of the origin, and if it is negative, then it's to the left of the origin.

Velocity has quite a similar interpretation. If a value is 0, it means we are at rest. Positive velocity means movement to the right, and negative velocity means movement to the left.

A 0 output for the acceleration function means velocity is either at a maximum, a minimum, or constant. Positive output means velocity is increasing, and negative output means velocity is decreasing.

#### Example:

Find the velocity if the displacement is given as a function of time .

We will simply differentiate this:

This is the same as [Look at differentiation if struggling]

v = 6t

If you wanted to find the acceleration, this could be written as the second derivative of displacement with respect to time or the velocity differentiated with respect to time.

a = $\frac{dv}{dt}=$6

So this particle has a Constant Acceleration of 6.

## Constant acceleration

Formally, this is known as 'one-dimensional Equations of motion for Constant Acceleration'. Kinematic equations in this section are only valid when acceleration is constant, and motion is constrained to a straight line.

As mentioned earlier, acceleration is constant, meaning it is the same regardless of time. We use standard Notation here, with a representing acceleration, u representing initial velocity, and t representing time.

### Motion graphs

Let's draw Graphs for all three properties of motion: displacement, velocity, and acceleration. We will identify the motion objects from the Graphs.

Acceleration time graph:

Here's the scenario: a bullet is fired vertically. The only force acting on it is gravity. Gravity doesn't change; therefore, the acceleration doesn't change either. Below is a demonstration of the acceleration time graph.

Acceleration

Velocity time graph:

Since we know that the acceleration is constant and negative because it is acting in the opposite direction to the velocity, that will mean the gradient of the velocity-time graph should be constant and negative too, which will give us something like this:

Velocity

Displacement time graph:

The graph below shows the bullet from its start point until it reaches maximum height, where the velocity is 0 before it starts coming down again. You have got a gradient that is constant and negative. So a graph for displacement will look like this:

Displacement

Note that velocity is always positive but decreases to 0. Therefore the gradient of the displacement vs time graph should always be positive and should decrease to 0.

### SUVAT equations

The SUVAT equations are a set of equations that describe motion in one dimension, each with one variable omitted.

 variable Description SI unit s Displacement meters (m) u Initial velocity m / s (meters per second) v Final velocity m / s (meters per second) a Acceleration m / s / s (meters per second per second) t Time s (seconds)

First equation: v = u + at

Second equation:

Third equation:

Forth equation:

#### Example:

A ferry carries passengers between the banks of a river which are 20m apart. After setting off, the ferry accelerates at 0.2 $m/{s}^{2}$ for 12 seconds before turning off the engine and decelerating at a constant rate and coming to a stop at the opposite bank.

1. Calculate the speed of the ferry after 12 seconds.

2. Calculate the distance the ferry travels during these 12 seconds.

1. Write your SUVAT down, clarify the variables you have and those you don't, and work out what equation to use.

s = x m

u = 0 m / s

v =?

a = 0.2 $m/{s}^{2}$

t = 12 s

Now we find the SUVAT equation which does not include s.

v = u + at

v = 0 + 0.2 $×$12

v = 2.4 m / s

1. Using

s = 14.4 m

## Projectiles

Projectiles deal with objects that are projected through the air either by being thrown, hit, fired, etc. Examples are:

• Someone throws a ball.

• A bullet at the instant it is fired.

• A car driving off a cliff.

A trajectory is the path of a projectile. A projectile always has some initial speed, usually angled to the horizontal. It moves freely under gravity because the only force acting on it is its weight. To solve a projectile problem, you need to use a SUVAT equation in two dimensions to split the horizontal and vertical components of each value.

Since Trigonometric Functions are used to find unknown Angles and distances, we will use that to resolve the components here.

#### Example:

A particle P is projected from a point O on a horizontal plane with a speed of 28m / s at 30 degrees to the horizontal. After projection, the particle moves freely under gravity until it strikes the plane at point A. Find the greatest height of the particle, the time of flight, and the distance OA.

We want to find a vertical height so we will use SUVAT in the vertical direction.

s =?

u = 28sin30 = 14 m / s

v = 0

a = -g $m/{s}^{2}$

t = x

With a situation like this, we use an equation that doesn't have t.

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

If we plug in the values, we will get 0 = 196 - 2gs

And we take g to be equal to 9.8, s = 10m

The second thing we have to find is the time of flight (the time it takes P to get from O to A).

s = 0 m

u = 14 m / s

v = x

a = -g $m/{s}^{2}$

t =?

$s=14t-\frac{g}{2}{t}^{2}=t\left(14-\frac{g}{2}t\right)$

So, t = 0 is when particle P is at the origin.

Whereas the solution where is when the particle is at A.

And if we take g = 9.8, t = 2.9s.

For the final part of the problem, we need to find the distance between O and A.

We are going to find a horizontal distance, so we will use SUVAT in the horizontal direction.

s =?

u = 28cos30 m / s

v = x

a = 0 $m/{s}^{2}$

The distance between O and A is s = 69 m

## Kinematics - key takeaways

• Displacement is the change in the position of an object relative to its frame of reference.
• Acceleration due to gravity is either 9.8 $m/{s}^{2}$ on the surface of the earth. It can be considered as 10 $m/{s}^{2}$ for the ease of calculation.
• You can use calculus techniques to analyze functions too.
• To solve kinematic problems in two dimensions, you need to use geometry to determine unknown magnitudes and directions and use the directions to determine kinematic quantities.

Kinematics is an area of study that focuses on the movement of objects.

Kinematics is used in mechanical engineering, robotics, and biomechanics to describe the motion of systems made of parts joined together.

## Test your knowledge with multiple choice flashcards

Which of these is an example of a projectile motion?

An object is moving in a straight line with a constant acceleration of 5 m/s². Which of the following curves express it on a displacement-time graph?

An object is moving in a straight line with a constant acceleration of 5 m/s². Which of the following curves express it on a velocity-time graph?

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