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Deriving the Suvat equations

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International Foundation Year (IFYSC0)

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Deriving the SUVAT equations

Foundation year physics notes, Dr Bianca Sala, 25 Aug 2020

We have already established in the slides that the total displacement covered is given by the area

under the velocity – time graph. Let’s first have a look at three different cases and derive expressions

for the displacement s from the velocity – time graphs.

When we have constant velocity i. no acceleration

The area under this graph is a rectangle, so to calculate the displacement we need to calculate the

area of the rectangle. This is just base × height.

𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = 𝑠 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑣 − 𝑡 𝑔𝑟𝑎𝑝ℎ = 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑡𝑖𝑚𝑒 × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑣𝑡

So if we have a constant velocity, the displacement will simply be velocity multiplied by time.

When we have constant acceleration and start with an initial velocity of zero

In this case we have a non - zero acceleration a, and our initial velocity u is zero i. we start from rest.

The final velocity reached is v.

In this case we need to calculate the area of the right angle triangle formed under the v – t graph. The

base of the triangle is time t and the height is the final velocity v. The area of a right angle triangle is a

half of the base multiplied by the height.

𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑠 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 =

𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 =

𝑡𝑖𝑚𝑒 × 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =

𝑡 × ∆𝑣 =

∆𝑣𝑡

The symbol Δ (delta) is used here to denote change, in this case the change in velocity Δv.

When we have constant acceleration, and start with an initial velocity u

In this case we have an initial velocity u and a final velocity v. The acceleration is still constant so our

v – t graph is a straight line.

It this case the area under the graph is made out of a rectangle and a triangle. We need to add these

two areas up when calculating the displacement. The height of the rectangle is u, while the height of

the triangle in this case is (u-v). The base is t in both cases.

𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑠 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 + 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 𝑢 × 𝑡 +

1

2 (𝑣 − 𝑢)𝑡

= 𝑢𝑡 +

1

2 𝑣𝑡 −

1

2 𝑢𝑡 =

1

2 𝑣𝑡 +

1

2 𝑢𝑡 =

1

2 (

𝑢 + 𝑣

)

𝑡

This is the most general equation for calculating the area under a v – t graph with constant acceleration

and will be our first SUVAT equation.

The first SUVAT equation

𝑠 =

1

2 (𝑢 + 𝑣)𝑡

From here we will focus on deriving the other SUVAT equations.

The second SUVAT equation

Let’s go back to our last graph, where the initial velocity was u and the final velocity was v. The graph

is a straight line, so we can use the straight line equation to describe it.

𝑆𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑦 = 𝑚𝑥 + 𝑐

Here, y is the velocity, m (the gradient) is the acceleration, x is the time and c (the y - intercept) is the

initial velocity u. So we can just rewrite the straight line equation as:

𝑣 = 𝑎𝑡 + 𝑢 = 𝑢 + 𝑎𝑡

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Deriving the Suvat equations

Module: International Foundation Year (IFYSC0)

6 Documents

Students shared 6 documents in this course

University: University of Sheffield

Was this document helpful?

Deriving the SUVAT equations

Foundation year physics notes, Dr Bianca Sala, 25 Aug 2020

We have already established in the slides that the total displacement covered is given by the area

under the velocity – time graph. Let’s first have a look at three different cases and derive expressions

for the displacement s from the velocity – time graphs.

1. When we have constant velocity i.e. no acceleration

The area under this graph is a rectangle, so to calculate the displacement we need to calculate the

area of the rectangle. This is just base

height.

𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 =𝑠=𝑎𝑟𝑒𝑎/𝑢𝑛𝑑𝑒𝑟/𝑣 − 𝑡/𝑔𝑟𝑎𝑝ℎ = 𝑏𝑎𝑠𝑒/ × ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑡𝑖𝑚𝑒/ × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =𝑣𝑡

So if we have a constant velocity, the displacement will simply be velocity multiplied by time.

2. When we have constant acceleration and start with an initial velocity of zero

In this case we have a non - zero acceleration a, and our initial velocity u is zero i.e. we start from rest.

The final velocity reached is v.

In this case we need to calculate the area of the right angle triangle formed under the v – t graph. The

base of the triangle is time t and the height is the final velocity v. The area of a right angle triangle is a

half of the base multiplied by the height.

Deriving the Suvat equations

International Foundation Year (IFYSC0)

University of Sheffield

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Deriving the SUVAT equations

𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = 𝑠 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑣 − 𝑡 𝑔𝑟𝑎𝑝ℎ = 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑡𝑖𝑚𝑒 × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑣𝑡

𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑠 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 =

𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 =

𝑡𝑖𝑚𝑒 × 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =

𝑡 × ∆𝑣 =

∆𝑣𝑡

𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑠 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 + 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 𝑢 × 𝑡 +

1

2

(𝑣 − 𝑢)𝑡

= 𝑢𝑡 +

1

2

𝑣𝑡 −

1

2

𝑢𝑡 =

1

2

𝑣𝑡 +

1

2

𝑢𝑡 =

1

2

(

𝑢 + 𝑣

)

𝑡

𝑠 =

1

2

(𝑢 + 𝑣)𝑡

𝑆𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑦 = 𝑚𝑥 + 𝑐

𝑣 = 𝑎𝑡 + 𝑢 = 𝑢 + 𝑎𝑡

Deriving the Suvat equations

Module: International Foundation Year (IFYSC0)

University: University of Sheffield

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