Kinematics formulae

In this chapter we shall be exploring the kinematics formulae. We shall be looking at selecting and applying the appropriate acceleration formulae, using vectors in problems involving constant acceleration. You must have some knowledge of vectors and acceleration graphs.

Deriving v=u+at

Here is the first kinematic formula. v=u+at Below we shall be looking at how this is formed. Here is the velocity-time graph of an object whose velocity changes with time.

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In the graph above;

u shows the initial velocity in ms^-1

v is the final velocity in ms^-1, at t seconds

We can use a to represent the acceleration in ms^-1 In this graph the acceleration is constant because of the straight line ie; the velocity is ever increasing. Acceleration is change of velocity over time.

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That is;

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We can transform the above to find the formula for velocity.

Multiply t in both sides of the equation;

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Now add u to both sides;

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Our first formula becomes;

We use S to represent displacement. We shall use S here to represent displacement of the object after t seconds. The standard unit measurement is metres represented with m

We can find the displacement by finding the area under the graph as shown below.

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The shape under the graph above is a trapezium. So we can conclude here that the displacement (S) is given by;

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This is generally written as;

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You must note that the letters S, u, v, a and t are the letters used in the acceleration formulae. As we’ve seen above for the first ones we’ve found;

u is the initial velocity in ms^-1

v is the velocity in ms^-1 after t seconds.

a is the acceleration in ms^-2.

S is the displacement.

Now let’s start again with the last formulae we found. The displacement formulae;

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In this formulae we can find another displacement formulae; Look at the above formulae. We have v and we already know the formula for v so we can replace v to get the equation;

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We simplify by collecting like terms to get;

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Simplify again and the equation becomes;

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Now we remove the brackets by expanding;

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The above formula is one of the kinematics formulae that we have found. That adds to the 3 formulae count that we have already found.

Let’s start again with one of the formulae we have already found. Let’s start with this one this time;

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Square both sides of the formula

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Now remove the brackets by expanding.

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If you look carefully you might realise that we might be able to form the S formula we found previous to form part of the above equation. There is a 2a factor which we can take out.

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Do you spot the S (displacement) formula. We replace this with S and out final derived formula becomes;

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So here are the kinematics formulae we’ve been able to find.

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Now let’s see how we can apply these formulae to different problems.

Example

An object has a constant acceleration with a speed of 7.8 ms^-1 After 4 seconds it speed is 9.2ms in a single direction. How far has the object travelled in this time?

First we identify what we have been given or know. We’re trying to find S (the displacement). We have u, v and t. The formula we need to use is;

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Substitute in the know values and then work out.

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After 4 seconds the object is 34m from its original position.

Example 2

A fast moving object passes point O with a speed of 20.1 ms^-1 and travels with a constant acceleration of 4.5 ms^-2 in the opposite direction. How far is the object from point O when it comes to rest?

Make it a good practice to list or identify the lettes; S, u, v, a and then identify which is know and which needs to be found. Another fact to note here is that a (acceleration) is negative since the problem specifies that the object is in the opposite direction.

These are the letters we have; s=?, u=20, v=0, a=-4.5. We need the formula which involves s, u, v and a, that must be;

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So now we substitute in the values and solve;

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The object is 44.4m from the point O when it comes to rest.

We could have used this following formula instead to find the time taken for the particle to come to rest.

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And then used the following formula to find the displacement;

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Example 3

An object has initial velocity of 12ms^-1 and a constant acceleration of -5.4ms^-2. What is the speed of the object after 7 seconds? After how long did the object come to rest?

Let’s identify what we have first. S=?, u=12, v=? a = -5.4ms^-2 and t=7.

Note here that the acceleration (a) is negative because the object is slowing down and the velocity is negative because the object is going in opposite direction. To find velocity we use the following formula;

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The speed of the object is 25.8ms^-1

The object has slowed down turned round and accelerated in the opposite direction at the speed of 25.8ms^-1 Since when it comes to rest the velocity will be zero we let the final velocity be zero in the following formula;

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The object came to rest after 2.2 seconds then went in the opposite direction.

Using vectors

In all the previous examples we have only looked at motion in a straight line. The formulas we found previously also work when u, v, a and S are vectors. The formulae were;

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Example

A particle P, initially position vector (,4 -7) m relative to a fixed point O. P starts with velocity -3 ms^-1 and moves with Acceleration (2i + 5j) ms^-2. Find the position vector of P after 6 seconds.

Let’s identify what we have s=?, v=-3i, v=?, a =2i+5j, t=6. The appropriate formula is;

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Now we substitute in what we know;

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This gives us the displacement from the original position. Now we need to add the displacement to the original position vector.

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The position vector of P after 6 seconds is (22i + 83j).

And that completes this chapter titled “Kinematics Formulae”. 😉