3.9 – Equations of Motion (SUVAT) | Physics with Mr Shone

Definition: The Four Equations of Motion
v = u + at s = ut + ½ at² v² = u² + 2as s = ½(u + v)t where: s : displacement (m) u : initial velocity (m/s) v : final velocity (m/s) a : acceleration (m/s²) t : time (s)

Definition: The Four Equations of Motion

v = u + at

s = ut + ½ at²

v² = u² + 2as

s = ½(u + v)t

where:

s : displacement (m)
u : initial velocity (m/s)
v : final velocity (m/s)
a : acceleration (m/s²)
t : time (s)

As the average velocity, <v>=½(u + v), we could write this last equation as s=<v>t

Notice that these equations all contain only four of the five SUVAT quantities. To determine Which equation we need, simply note down which values we have been given and which we need to find.

Note: These equations are only applicable for motion with uniform acceleration.

Mathematical derivation of the equations (You do not need to know or memorise this)

Worked Example 1
A car speeds up from rest on a straight road. When the car passes by point A, its velocity is 10 m s^-1. When the car passes point B, its velocity increases to 30 m s^-1. The car takes 5.0 seconds to travel from point A to point B. Calculate the distance travelled by the car in the 5.0 seconds. 1. Sketch a diagram: This will help you to visualise the situation. In particular it will be useful to determine a sign convention. 2. List all variables: u: 10 m s^-1 v: 30 m s^-1t: 5.0 s s: ? Note: We are looking at the motion from A to B, thus the initial velocity was not taken to be at rest. Putting down the unknown helps to identify which of the equations of motion we should use. 3. Write down equation: As we have u, v, t and s we must be using: s = ½(u + v)t 4. Substitute in values and solve: s = ½(10 + 30)5.0 s = 100 m

Worked Example 1

A car speeds up from rest on a straight road. When the car passes by point A, its velocity is 10 m s^-1. When the car passes point B, its velocity increases to 30 m s^-1. The car takes 5.0 seconds to travel from point A to point B.
Calculate the distance travelled by the car in the 5.0 seconds.

1. Sketch a diagram:
This will help you to visualise the situation. In particular it will be useful to determine a sign convention.

2. List all variables:
u: 10 m s^-1
v: 30 m s^-1t: 5.0 s
s: ?

Note: We are looking at the motion from A to B, thus the initial velocity was not taken to be at rest. Putting down the unknown helps to identify which of the equations of motion we should use.

3. Write down equation:
As we have u, v, t and s we must be using:
s = ½(u + v)t

4. Substitute in values and solve:
s = ½(10 + 30)5.0

s = 100 m

Worked Example 2
A ball is released from a height above the ground. The ball hits the ground 4.0 s after being released. Calculate the height it was released from. 1. Sketch a diagram: 2. List all variables: u: 0 m/s (released = dropped = no force parted to ball) t: 4.0 s a: 10.0 m s^-2s: ? 3. Write down equation: s = ut + ½ at² 4. Substitute in values and solve: s = 0×4.0 + ½×10×4² s = 80 m

Worked Example 2

A ball is released from a height above the ground. The ball hits the ground 4.0 s after being released. Calculate the height it was released from.

1. Sketch a diagram:

2. List all variables:
u: 0 m/s (released = dropped = no force parted to ball)
t: 4.0 s
a: 10.0 m s^-2s: ?

3. Write down equation:
s = ut + ½ at²

4. Substitute in values and solve:
s = 0×4.0 + ½×10×4²

s = 80 m

Worked Example 3
A ball is thrown upwards with a velocity of 40 m s^-1. Calculate the maximum height attainable by the ball, assuming that g is 10 m s^-2. u: 40 m s^-1 (sign convention is thus upwards is positive) v: 0 m s^-1 a: −10.0 m s^-2s: ? v² = u² + 2as 0 = 40²+ 2 × (−10) × s s = 80 m

Worked Example 3

A ball is thrown upwards with a velocity of 40 m s^-1. Calculate the maximum height attainable by the ball, assuming that g is 10 m s^-2.

u: 40 m s^-1 (sign convention is thus upwards is positive)
v: 0 m s^-1
a: −10.0 m s^-2s: ?

v² = u² + 2as
0 = 40²+ 2 × (−10) × s

s = 80 m

Worked Example 4
A ball is thrown upwards with a velocity of 40 m s^-1 at the top of a 100 m high cliff. Determine (a) the time the ball takes to reach the ground, and Taking upwards as positive: u: 40 m s¯¹ s: −100 m a: −10 m s¯² t: ? s = ut + ½ at² −100 = 40 × t + ½ (−10)×t² −100 = 40t−5t² −20 = 8t−t² t= 10.0 s Note: of course, solving a quadratic we get a second answer (t=−2.00 s), but we ignore this answer as a negative time make no sense. (b) the velocity at which it hits the ground. Taking upwards as positive: u: 40 m s¯¹ s: −100 m a: −10 m s¯² v: ? v² = u² + 2as v² = 40² + 2(−10)×(−100) v² = 1600 + 2000 v =√3600 v= −60 m/s Note: we need to be consistent with the sign convention we are using (upwards being positive in this example). As the ball is very obviously moving down when it hits the ground we take the negative value.

Worked Example 4

A ball is thrown upwards with a velocity of 40 m s^-1 at the top of a 100 m high cliff. Determine
(a) the time the ball takes to reach the ground, and

Taking upwards as positive:
u: 40 m s¯¹
s: −100 m
a: −10 m s¯²
t: ?

s = ut + ½ at²

−100 = 40 × t + ½ (−10)×t²

−100 = 40t−5t²

−20 = 8t−t²

t= 10.0 s

Note: of course, solving a quadratic we get a second answer (t=−2.00 s), but we ignore this answer as a negative time make no sense.

(b) the velocity at which it hits the ground.

Taking upwards as positive:
u: 40 m s¯¹
s: −100 m
a: −10 m s¯²
v: ?

v² = u² + 2as
v² = 40² + 2(−10)×(−100)
v² = 1600 + 2000
v =√3600

v= −60 m/s

Note: we need to be consistent with the sign convention we are using (upwards being positive in this example). As the ball is very obviously moving down when it hits the ground we take the negative value.

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