Uniform Acceleration (DP IB Physics: SL)

Deriving Kinematic Equations of Motion

• The kinematic equations of motion are a set of four equations that can describe any object moving with constant acceleration
• They relate the five variables:
  • s = displacement
  • u = initial velocity
  • v = final velocity
  • a = acceleration
  • t = time interval

• Knowing where these equations come from and how they are derived helps to understand them:

Derivation of

The average velocity is halfway between u and v

The two terms ut and ½at² make up the area under the graph

This final equation can be derived from two of the others

Summary of the four equations of uniformly accelerated motion

Key Takeaways

• These are all given in the IB DP Physics data booklet
• The key terms to look out for are:

'Starts from rest',

• This means u = 0 and t = 0
• This can also be assumed if the initial velocity is not mentioned

'Falling due to gravity'

• This means a = g = 9.81 m s^–2
  • It doesn't matter which way is positive or negative for the scenario, as long as it is consistent for all the vector quantities

• For example, if downwards is negative then for a ball travelling upwards, s must be positive and a must be negative

'Constant acceleration in a straight line'

• This is a key indication SUVAT equations are intended to be used
  • For example, an object falling in a uniform gravitational field without air resistance

How to use the SUVAT equations

• Step 1: Write out the variables that are given in the question, both known and unknown, and use the context of the question to deduce any quantities that aren’t explicitly given
  • e.g. for vertical motion a = ± 9.81 m s^–2, an object which starts or finishes at rest will have u = 0 or v = 0

• Step 2: Choose the equation which contains the quantities you have listed
  • e.g. the equation that links s, u, a and t is s = ut + ½at²

• Step 3: Convert any units to SI units and then insert the quantities into the equation and rearrange algebraically to determine the answer

Part (a)

Step 1: List the known quantities

• • • Initial velocity, u = 6 m s^–1 East
    • Acceleration, a = 2 m s^–2  East
    • Time, t = 4 s
    • Displacement, s = ? (this needs to be calculated)

Step 2: Identify and write down the equation to use

• • • Since the question states constant acceleration, SUVAT equations can be used
    • In this problem, the equation that links s, u, a, and t is

s = (u × t) + (½ × a × t²)

Step 3: Substitute known quantities into the equation and simplify where possible

s = (6 × 4) + (0.5 × 2 × 4²)

• • • This can be simplified to:

s = 24 + 16 = 40 m

Part (b)

Step 1: List the known quantities

• • • Initial velocity, u = 6 m s^–1 East
    • Acceleration, a = 2 m s^–2  East
    • Time, t = 4 s
    • final velocity, v = ? (this needs to be calculated)

Step 2: Identify and write down the equation to use

• • • Since the question states constant acceleration - SUVAT equation(s) - can be used
    • In this problem, the equation that links v, u, a, and t is v = u + (a × t)

Step 3: Substitute known quantities into the equation and simplify where possible

v = 6 + (2 × 4)

• • • This can be simplified to:

v = 14 m s^–1

Part (c)

Step 1: List the known quantities for cyclist A

• • • Initial velocity, u = 18 m s^–1 East
    • Acceleration, a = 0 m s^–2  East
    • Final velocity, v = 18 m s^–1 East
    • Time, t = ?
    • Displacement, s = ?

Step 2: List the known quantities for cyclist B

• • • Initial velocity, u = 0 m s^–1 East
    • Acceleration, a = 1.5 m s^–2  East
    • Final velocity, v = ?
    • Time, t = ?
    • Displacement, s = ?

Step 3: Express the situation for cyclist A and B in terms of displacement, s

• • • Cyclist A can have their situation expressed by:

s_A = (u × t) + (½ × a × t²)

s_A = (18 × t) + (½ × 0 × t²) = (18 × t)

• • • Cyclist B can have their situation expressed by:

s_B = (u × t) + (½ × a × t²)

s_B = (0 × t) + (½ × 1.5 × t²) = (0.75 × t²)

Step 4: Equate the two equations and solve for t

• • • The two equations describe the displacement of each cyclist respectively
    • When equating them, this will find the time when the cyclists are at the same location

s_A= s_B = (18 × t) = (0.75 × t²)

0 = (0.75 × t²) – (18 × t)

0 = (0.75 × t – 18) × t

• • • Therefore, solving for t, it can be two possible answers:

t = 0 s or 18 / 0.75 = 24 s

Step 4: State the final answer

• • • Since the question is seeking the time when the two cyclists meet after first passing each other, the final answer is 24 s