How to Solve: Kinematics (SUVAT Equations, Motion Graphs, Variable Acceleration with Calculus)

How to Solve: Kinematics (SUVAT Equations, Motion Graphs, Variable Acceleration with Calculus)

Introduction

"Mastering kinematics unlocks 10–15% of your A-Level Mechanics paper—and it’s the difference between a C and an A. Whether it’s a rocket launch, a car braking, or a ball thrown off a cliff, these equations predict real-world motion. Let’s break it down so you can solve any problem in under 2 minutes."

What You Need To Know First

1. Basic algebra – Rearranging equations, solving for unknowns.
2. Graphs – Understanding distance-time and velocity-time graphs.
3. Calculus basics (A-Level only) – Differentiating and integrating simple functions (e.g., ( t^2 ), ( t^3 )).

Key Vocabulary

Formulas To Know

SUVAT Equations (MEMORISE THESE)

1. ( v = u + at )
2. ( v ) = final velocity (m/s)
3. ( u ) = initial velocity (m/s)
4. ( a ) = acceleration (m/s²)

5. ( t ) = time (s)

6. ( s = ut + \frac{1}{2}at^2 )

7. ( s ) = displacement (m)

8. ( v^2 = u^2 + 2as )

9. ( s = \frac{1}{2}(u + v)t )

10. ( s = vt - \frac{1}{2}at^2 )

Motion Graphs (GIVEN ON EXAM SHEET, BUT KNOW HOW TO USE)

• Gradient of ( s-t ) graph = velocity.
• Gradient of ( v-t ) graph = acceleration.
• Area under ( v-t ) graph = displacement.

Variable Acceleration (A-Level Only – MEMORISE THESE)

• If ( v = f(t) ), then:
• Acceleration ( a = \frac{dv}{dt} ).
• Displacement ( s = \int v \, dt ).
• If ( a = f(t) ), then:
• Velocity ( v = \int a \, dt ).
• Displacement ( s = \int v \, dt ).

Step-by-Step Method

For SUVAT Problems:

1. Read the question. Underline the given values and what you need to find.
2. Choose the right SUVAT equation. Pick the one that uses the variables you have and the one you need.
3. Convert units if needed. (e.g., km/h → m/s: divide by 3.6).
4. Plug in the numbers. Write the equation with the known values.
5. Solve for the unknown. Rearrange and calculate.
6. Check your answer. Does it make sense? (e.g., negative displacement = wrong direction).

For Motion Graphs:

1. Identify the graph type. Is it ( s-t ), ( v-t ), or ( a-t )?
2. Find the gradient (for velocity/acceleration) or area under the graph (for displacement).
3. Break the graph into shapes (triangles, rectangles) if needed.
4. Calculate the area/gradient for each section.
5. Add/subtract to find total displacement or average velocity.

For Variable Acceleration (A-Level):

1. Write the given function. (e.g., ( v = 3t^2 + 2t )).
2. Differentiate for acceleration or integrate for displacement.
3. Add constants of integration if initial conditions are given.
4. Solve for the unknown (e.g., find ( s ) at ( t = 5 )).
5. Check units and reasonableness.

Worked Examples

Example 1 – Basic SUVAT

Question: A car accelerates from rest at ( 2 \, m/s^2 ) for 5 seconds. What is its final velocity?

Solution: 1. Given:
- ( u = 0 \, m/s ) (starts from rest)
- ( a = 2 \, m/s^2 )
- ( t = 5 \, s )
- Find: ( v )

1. Choose equation: ( v = u + at ) (uses ( u, a, t ), finds ( v )).

2. Plug in:
   ( v = 0 + (2)(5) )

3. Calculate:
   ( v = 10 \, m/s )

What we did and why: We used the simplest SUVAT equation because we had ( u, a, t ) and needed ( v ). Always start with the equation that uses the most given values.

Example 2 – Medium SUVAT (Missing Time)

Question: A ball is thrown upwards at ( 15 \, m/s ). How high does it go? (Take ( g = 9.8 \, m/s^2 ) downwards.)

Solution: 1. Given:
- ( u = 15 \, m/s ) (upwards)
- ( v = 0 \, m/s ) (at max height, velocity is 0)
- ( a = -9.8 \, m/s^2 ) (acceleration is downwards)
- Find: ( s )

1. Choose equation: ( v^2 = u^2 + 2as ) (uses ( u, v, a ), finds ( s )).

2. Plug in:
   ( 0 = (15)^2 + 2(-9.8)s )

3. Solve for ( s ):
   ( 0 = 225 - 19.6s )
   ( 19.6s = 225 )
   ( s = \frac{225}{19.6} \approx 11.48 \, m )

What we did and why: We used ( v^2 = u^2 + 2as ) because time wasn’t given. The negative acceleration accounts for gravity acting downwards.

Example 3 – Exam-Style (Motion Graphs)

Question: The velocity-time graph of a cyclist is shown below. Calculate the total displacement.

(Graph description: 0–4s: straight line from 0 to 8 m/s; 4–10s: horizontal line at 8 m/s.)

Solution: 1. Break the graph into shapes:
- 0–4s: Triangle (area = ( \frac{1}{2} \times base \times height ))
- 4–10s: Rectangle (area = ( base \times height ))

1. Calculate areas:
2. Triangle: ( \frac{1}{2} \times 4 \times 8 = 16 \, m )

3. Rectangle: ( 6 \times 8 = 48 \, m )

4. Total displacement: ( 16 + 48 = 64 \, m )

What we did and why: We split the graph into simple shapes to find the area under the curve, which gives displacement. Always check if the graph is above/below the x-axis (negative area = opposite direction).

Example 4 – Variable Acceleration (A-Level)

Question: The velocity of a particle is given by ( v = 4t^2 + 3t ). Find its displacement after 2 seconds.

Solution: 1. Given: ( v = 4t^2 + 3t )
- Find: ( s ) at ( t = 2 )

1. Integrate ( v ) to find ( s ):
   ( s = \int v \, dt = \int (4t^2 + 3t) \, dt )
   ( s = \frac{4t^3}{3} + \frac{3t^2}{2} + C )

2. Assume ( s = 0 ) at ( t = 0 ), so ( C = 0 ).

3. Plug in ( t = 2 ):
   ( s = \frac{4(2)^3}{3} + \frac{3(2)^2}{2} )
   ( s = \frac{32}{3} + 6 )
   ( s = \frac{32}{3} + \frac{18}{3} = \frac{50}{3} \approx 16.67 \, m )

What we did and why: We integrated the velocity function to find displacement. The constant ( C ) was zero because the particle started at the origin.

Common Mistakes

Exam Traps

1-Minute Recap

"Here’s the night-before cheat sheet: 1. SUVAT: Write down the 5 equations. Pick the one that uses the most given values. 2. Graphs: Gradient = velocity/acceleration. Area = displacement. 3. Variable acceleration: Differentiate for acceleration, integrate for displacement. 4. Signs matter! Up = +, down = –. Always check units. 5. Practice 3 problems tonight: One SUVAT, one graph, one calculus (A-Level). You’ve got this!