Equations of motion

Higher Physics

• Experiments show that at a particular place all bodies falling freely under gravity, in a vacuum or where air resistance is negligible, have the same constant acceleration irrespective of their masses.

• This acceleration towards the surface of the Earth, known as the acceleration due to gravity, is donated by g.

• Its magnitude varies slightly from place to place on the Earth´s surface and is approximately 9.8ms-2

Acceleration Due to Gravity

The Effects of Air Resistance

• Air resistance depends on 2 things– Surface area– Velocity

• Air resistance increases as surface area increases

• Air resistance increases as the velocity increases

Terminal Velocity

• As an object falls through the air, it accelerates, due to the force of attraction of the Earth. This force does not change.

• As the velocity increases, the air resistance, the force opposing the motion, increases, therefore the acceleration decreases.

• If the object falls for long enough, then the air resistance (a force acting upwards) will equal the force of attraction of the Earth (the weight) (a force acting downwards)

• Now there are no net forces acting on the object (since the two forces balance) so it no longer accelerates, but travels at a constant velocity called its terminal velocity.

• Terminal velocity depends on– The size

– Shape

– And weight of the object

• A sky diver has a terminal velocity of more than 50ms-1 (how many km per hour?)

Equations of motion

• Linear motion symbols:

• u - initial velocity (m/s)

• v - final velocity (m/s)

• a - acceleration (m/s2)

• s - displacement (m)

• t - time taken (s)

Equations of motion

• There are three equations of motion:

atuv 2

2

1atuts

asuv 222

•The equations of motion can be used for projectiles and objects moving in a straight line (cars etc).

Equations of motion

• Example 1:A car is travelling with a velocity of 5 m/s. It then accelerates uniformly and travels a distance of 50m. The velocity reached is 15 m/s.

• (a) Find the acceleration of the car.• (b) Calculate the time taken to travel this

distance.

Solution

• 1. Write down what you know:

• 2. S= 50m, u = 5m/s , v=15m/s , a=a

• 3. As you are trying to find ‘a’ just write the letter ‘a’

• 4. Select the equation to use.

Solution

2

22

22

/2100

200

10025225

502515

2

sma

a

a

asuv

•(a) Find the acceleration of the car.

Solution

st

t

t

atuv

5

102

2515

(b) Calculate the time taken to travel this distance.

Projectiles

• A projectile is an object projected by force and continuing in motion by its own inertia. (It is not moving under the effects of an engine)

Projectile motion

• A projectile’s motion can be simplified by considering the horizontal and vertical components separately. These are independent of each other.

• The subscript v is used for vertical motion and h is used for horizontal motion.

• A projectile follows the path shown below.

• The horizontal component of velocity is constant throughout the projectiles motion.

Path of a Projectile

A

B

CD

E

• Thus Vh = Uh

Video - Constant Horizontal Motion

• Watch the video. How does this show constant horizontal velocity?

• How do the initial horizontal velocities of the plane and bomb compare?

Path of projectile

• The vertical component of the velocity changes throughout due to gravity.• A - Vertical velocity is maximum.• B - Vertical velocity decreases due to effect of gravity. • C - Maximum height, the vertical velocity is zero.

A

B

CD

E

Path of projectile

A

B

CD

E

•D - the projectiles velocity then starts to increase due to the acceleration of gravity. The velocity will now be negative.

•E - the projectile lands with the same velocity that it started with but is negative in magnitude.

•For projectile problems assume that upwards is positive and downwards is negative.

Dealing with projectile problems

• Use this format to record data available.Horizontal Verticaluh =t =sh =

uv=vv=av= -9.8 ms-2

sv=t =

tus hh Use equations of motion

Example 2

• A golfer hits a ball with a velocity of 48ms-1 at an angle of 300 as shown below.

• (a) What is the horizontal and vertical components of velocity?.

• (b) What is the resultant velocity of the ball after 2 seconds.

48 ms-1

300

Solution

• (a) Update diagram

• uh = 48 cos 30 = 41.56 ms-1

• uv = 48 sin 30 = 24 ms-1

48 ms-1

300

uh

uv

• (b) Horizontal component is constant , so final velocity (vh) = 41.56ms-1

• Vertical velocity has changed , use format and equations of motion to find vv .

uv= 24ms-1

vv= ???av= -9.8 ms-2

sv=t = 2s

14.4

)28.9(24

msv

v

atuv

v

v

vv

• (b) To complete question , calculate the resultant velocity.

• Use pythagoras to find resultant velocity

?? ms-1

vh= 41.56ms-1

vv= 4.4 ms-1

Example 3

• A stone is dropped down a water well and it hits the surface of the water after 4s.

• What was the velocity of the stone when it hit the water?uv= 0ms-1

vv= ???av= -9.8 ms-2

sv=t = 4s

12.39

)48.9(0

msv

v

atuv

v

v

vv

Example 4

• An object is thrown horizontally from a cliff and takes 6s to land. If it lands 60m away from the base of the cliff, find:

• (a) its horizontal velocity

• (b) the height of the cliff

• (c) its velocity (size and direction) on impact.

Example 5

• A balloon is climbing vertically with a velocity of 5ms-1 when a sandbag is dropped from it and hits the ground 3 seconds later.

• Find• (a) The velocity of the sandbag as it hits the

ground.• (b) The original height of the sandbag before

it is dropped.