**SUVAT Equations**

You need to learn the equations below

**s = displacement**(distance with direction - a particle can travel 30m but still have a displacement of 0m if it returns back to its starting position)

**u = starting velocity**(speed with direction)

**v = final velocity**(speed with direction - can have minus velocity)

**a = acceleration**(also has direction - deceleration is acceleration in the opposite direction to the direction of travel)

**t = time**

In mechanics the

**signs**(+ or -) need to be viewed as

**direction**not as size

**Positive**direction is generally accepted to be

**right**up the x axis and

**up**the y axis

**Negative**direction is generally accepted to be

**left**down the x axis and

**down**the y axis

For example:

**Acceleration in Straight Line Kinetics**

**Unless told otherwise**a particle will

**continue at the same acceleration**

If a particle is

**accelerating**in the

**negative direction**but traveling with a

**positive velocity**the particle will

**decelerate**

However once the

**velocity reaches zero**the particle is

**still accelerating**in the negative direction so it will then start traveling in the

**negative direction**

For particle P from O to T:

To find out the

**s = ?m**

u = 8m/s

v = 0m/s

a = -2m/s^2

t = ?su = 8m/s

v = 0m/s

a = -2m/s^2

t = ?s

To find out the

**maximum displacement**(achieved at T when the particle travels no further):Because the rate of change in velocity (acceleration) is the same at all times we can assume things are the same on the journey back to

From

From

This means the displacement of particle P from

In this case

What's different is the time.

To work out the time taken to get from

*O*.From

*O*to*T*= 16mFrom

*T*to*O*= -16mThis means the displacement of particle P from

*O*back to*O*is 0mIn this case

**every point**from*O*to*T*the particle has**two velocities**of**equal magnitude**but**opposite direction**What's different is the time.

To work out the time taken to get from

*O*to*O*use:**s = 0m****u = 8m/s****v = -8m/s****a = -2m/s^2****t = ?s**From being told that a particle started at

If given the displacement of

*O*traveling at**8 m/s**decelerating at**2 m/s^2**we have been able to deduce that the:**Maximum displacement**from*O*in positive direction is**16m**- The
**velocity**at pointcan be**O****8 m/s**and**-8 m/s**at different times (t = 0 and t = 8) - The
**time**it takes for particle P to**return**to*O*is**8s**

If given the displacement of

*A*or the time the particle arrives at*A*we can deduce the velocity at*A***Acceleration and Gravity**

**Gravity makes an object accelerate as it is a**

__force__Gravity provides a

**downward**acceleration of

**9.8 m/s^2**

In the vertical plain works in exactly the same way as in a horizontal line except we don't need to be told as much.

*S*to*T*is 5m and*S*to*G*takes 0.2sWe need to find out what the

**starting velocity**of P is at

*S*

At first look we've only been told one thing about

*T*. However we know much more:

**s = 5m**

**u = ?m/s**

**v = 0m/s**(because T is the turning point)

**a = -9.8m/s^2**(due to gravity)

**t = ?s**

To find out 'u' use:

To find the height P started off the ground use:

**s = ?m**

u = -9.9m/s

v = ?m/s

a = -9.8m/s^2

t = 0.2su = -9.9m/s

v = ?m/s

a = -9.8m/s^2

t = 0.2s

Use:

This means the particle started 2.2m above

*G***Two particles**

When asked to work out when two particles match each other's displacement, velocity etc. you need to use simultaneous equations.

E.g.

Particle P is traveling at a constant velocity of 5m/s

3 seconds later Particle Q sets off from the same place and accelerates at a constant rate of 3m/s^2

How long after Q starts does Q overtake P?

First right down the information you know that starts from

P

Q

This means you want an equation without v in it

E.g.

Particle P is traveling at a constant velocity of 5m/s

3 seconds later Particle Q sets off from the same place and accelerates at a constant rate of 3m/s^2

How long after Q starts does Q overtake P?

First right down the information you know that starts from

**t = 3**P

**s = ?**

u = 5

v = 5

a = 0

t = ?u = 5

v = 5

a = 0

t = ?

Q

**s = ?**

u = 0

v = ?

a = 3

t = ?

u = 0

v = ?

a = 3

t = ?

This means you want an equation without v in it

**Kinetic Graphs**

*Graphs are a good way of working out questions with two particles. This helps you plot where they meet each other.*