# Differentiation with Linear Kinematics

## Differentiation with Linear Kinematics Revision

**Differentiation with Linear Kinematics**

**Kinematics **looks at the motion of a particle, considering **displacement**, **velocity**, and **acceleration**. We can model the movement of anything as a particle, such as a car or a ball.

Make sure you are happy with the following topic before continuing:

**Displacement**

**Displacement** is slightly different to distance…

We know distance is a measurement that is always positive – you cannot get a negative distance.

**Displacement** however, measures the distance from a fixed point of origin, and if this distance is forwards from the origin, the** displacement** is positive, and if it is backwards, the **displacement** is negative. We call **displacement** s, and it is usually measured in \text{m}.

For example, both the distance and **displacement** of the following ball from the origin is 4:

However, in the following example, while the distance is 3, the **displacement** is -3:

**Velocity**

**Velocity** measures a change of **displacement**, or how quickly the **displacement** is changing.

**Velocity** is usually written v, and measured in \text{m/s}. We can look at it in a similar way to **displacement**:

**Velocity** is slightly different to speed, as it also takes direction into consideration. If v = 4, then the particle is moving in the positive direction with a speed of 4 \text{m/s}. If v = -4, then the particle is moving in the negative direction with a speed of 4 \text{m/s}.

**Acceleration**

**Acceleration** measures a change of **velocity**, or how quickly the **velocity** is changing.

**Acceleration** is usually written a, and measured in \text{m/s}^2.

If a = 3, the particle is accelerating at 3\text{ m/s}^2.

If a = -3, the particle is decelerating at 3\text{ m/s}^2.

**Using Differentiation**

We can use differentiation to help us find **displacement**, **velocity**, and **acceleration**…

We express **displacement**,** velocity**, and **acceleration** in terms of t, which is time, usually in seconds.

If we differentiate an expression for **displacement**, we would get an expression for **velocity**, and if we differentiate an expression for **velocity**, we would get an expression for **acceleration**.

**Example 1: Using Differentiation**

The **displacement** of a car is shown in the following expression:

s = 6t^3 + 5t^2 + 11t -4

Find an expression for the **velocity** and **acceleration** of this car.

**[4 marks]**

To find **velocity**, we differentiate once:

s = 6t^3 + 5t^2 + 11t -4\\

\dfrac{\text{d}s}{\text{d}t} = 18t^2 + 10t + 11

To find **acceleration**, we differentiate the **velocity**:

v = 18t^2 + 10t + 11\\

\dfrac{\text{d}v}{\text{d}t} = 36t + 10

**Example 2: Applied Differentiation**

A ball is moving along a straight line. The **displacement** of the particle, from fixed origin point O, at time, t seconds is:

s = 7t^2 - 5t - 12

**a)** Find the particle’s initial distance from O

**b)** Find t when the particle is at rest

**[5 marks]**

**a)** ‘Initial’ means t=0, so we can use t=0 in the **displacement** equation to find the initial distance:

s = 7t^2 - 5t - 12\\

s = 7(0)^2 - 5(0) - 12\\

s = -12

The initial **displacement** is -12, however, the question is asking for distance, so this is 12\text{ m}.

**b)** When the particle is at rest, the **velocity** will be 0\text{ m/s}^2, so we need to firstly differentiate to find the expression for **velocity**.

s = 7t^2 - 5t - 12\\

\dfrac{\text{d}s}{\text{d}t} = 14t - 5\\

Now we can sub in \dfrac{ds}{dt}=0\text{ m/s}^2

0 = 14t - 5\\

14t = 5\\

t = \dfrac{5}{14}\text{ s}

## Differentiation with Linear Kinematics Example Questions

**Question 1: **The velocity of a particle is given by the expression

v = 21t^2 + 5t - 11

Find an expression for the acceleration of the particle.

**[2 marks]**

We need to differentiate the velocity to find the acceleration:

v = 21t^2 + 5t - 11\\

\dfrac{\text{d}v}{\text{d}t} = 42t + 5

**Question 1: **The displacement of a train is given by the expression

s = 7t^2 - 14t + 6

Find the velocity and acceleration of the train at t = 2\text{ s}.

**[4 marks]**

Differentiate to find an expression for velocity:

s = 7t^2 - 14t + 6\\

\dfrac{\text{d}s}{\text{d}t} = 14t - 14

When t = 2\text{ s}, the velocity is:

v = 14(2) - 14\\

v = 14\text{ m/s}

Differentiate to find an expression for acceleration:

v = 14t - 14\\

\dfrac{\text{d}v}{\text{d}t} = 14

At t=2\text{ s}, the acceleration is 14\text{ m/s}^2

**Question 3: **The displacement of a particle is given by the expression

s = 3t^2 + 12t - 2

Work out whether the acceleration of this particle remains constant or is variable with time.

**[4 marks]**

Let’s find an expression for the acceleration:

s = 3t^2 + 12t - 2\\

\dfrac{\text{d}s}{\text{d}t} = 6t+12\\

\dfrac{\text{d}v}{\text{d}t} = 6

So, the acceleration is 6\text{ m/s}^2.

As there is no t in the equation for acceleration, we know it will not change if t changes, and so will always remain constant at 6\text{ m/s}^2.