GCSE — physicsMedium6 min read

Velocity and Acceleration

Velocity-time graphs; calculating acceleration; area under v-t graph = distance

Tutor AI Team
Tutor AI Team

# Velocity and Acceleration — GCSE Physics

Acceleration is one of the most important concepts in physics — it describes how quickly an object's velocity changes. Combined with velocity-time graphs, it gives us powerful tools to analyse the motion of everything from cars to rockets.

1. Velocity

Velocity is the speed of an object in a given direction. It is a vector quantity.

velocity=displacementtime\text{velocity} = \frac{\text{displacement}}{\text{time}}

  • Displacement is the distance in a specific direction (vector)
  • Distance is how far an object has moved regardless of direction (scalar)

Example: If you walk 100 m north and then 100 m south, your:

  • Distance = 200 m
  • Displacement = 0 m (you're back where you started)

2. Acceleration

Acceleration is the rate of change of velocity. It tells you how quickly the velocity is changing.

a=ΔvΔt=v−ut\boxed{a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}}

Where:

  • aa = acceleration (m/s²)
  • vv = final velocity (m/s)
  • uu = initial velocity (m/s)
  • tt = time taken (s)
  • Δv\Delta v = change in velocity

Understanding the Units

m/s² means "metres per second, per second." An acceleration of 3 m/s² means the velocity increases by 3 m/s every second.

Positive and Negative Acceleration

Scenario Acceleration Common Name
Speed increasing Positive Acceleration
Speed decreasing Negative Deceleration
Constant speed Zero No acceleration
Direction changing (constant speed) Non-zero Centripetal acceleration

Deceleration is a negative acceleration — the object is slowing down.

Rearranging

v=u+att=v−uav = u + at \qquad t = \frac{v - u}{a}

3. Uniform vs Non-Uniform Acceleration

Uniform (constant) acceleration: The velocity changes by the same amount each second.

  • Example: A ball falling freely near Earth's surface accelerates at 9.89.8 m/s² uniformly.

Non-uniform acceleration: The rate of velocity change varies over time.

  • Example: A car accelerating in real traffic — the driver adjusts the throttle.

4. Velocity-Time Graphs

A velocity-time (v-t) graph shows how velocity changes over time.

Reading v-t Graphs

Feature Meaning
Horizontal line Constant velocity (zero acceleration)
Straight line sloping upward Constant (uniform) acceleration
Straight line sloping downward Constant (uniform) deceleration
Steeper slope Greater acceleration
Curve Non-uniform acceleration
Line on x-axis (v = 0) Object is stationary
Line below x-axis Moving in the opposite direction

The Gradient = Acceleration

Acceleration=gradient=change in velocitychange in time=ΔvΔt\text{Acceleration} = \text{gradient} = \frac{\text{change in velocity}}{\text{change in time}} = \frac{\Delta v}{\Delta t}

  • Positive gradient = acceleration
  • Negative gradient = deceleration
  • Zero gradient (horizontal) = constant velocity

The Area Under the Graph = Distance

The area between the line and the time axis on a v-t graph gives the distance travelled (or displacement).

For simple shapes:

  • Rectangle: area = base × height = t×vt \times v
  • Triangle: area = 12×base×height=12×t×v\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times v
  • Trapezium: area = 12(a+b)×h\frac{1}{2}(a + b) \times h

5. SUVAT Equations (Higher Tier)

For uniform acceleration problems, these equations connect the five variables:

Symbol Meaning
ss displacement
uu initial velocity
vv final velocity
aa acceleration
tt time

The key equations:

v=u+atv = u + at s=ut+12at2s = ut + \frac{1}{2}at^2 v2=u2+2asv^2 = u^2 + 2as s=(u+v)2×ts = \frac{(u + v)}{2} \times t

To solve a problem:

  1. List what you know (ss, uu, vv, aa, tt)
  2. Identify what you need to find
  3. Choose the equation that contains the known quantities and the unknown
  4. Substitute and solve

Worked Example: Calculating Acceleration

Problem

Question: A car accelerates from 10 m/s to 30 m/s in 5 seconds. Calculate the acceleration.

Solution

a=v−ut=30−105=205=4 m/s2a = \frac{v - u}{t} = \frac{30 - 10}{5} = \frac{20}{5} = 4 \text{ m/s}^2

Worked Example: Calculating Deceleration

Problem

Question: A bus travelling at 18 m/s brakes and stops in 6 seconds. Calculate the deceleration.

Solution

a=v−ut=0−186=−186=−3 m/s2a = \frac{v - u}{t} = \frac{0 - 18}{6} = \frac{-18}{6} = -3 \text{ m/s}^2

Deceleration = 3 m/s².

Worked Example: Distance from v-t Graph

Problem

Question: A v-t graph shows a car accelerating uniformly from 0 to 20 m/s in 10 seconds. Calculate the distance travelled.

Solution

The shape under the graph is a triangle: Distance=12×base×height=12×10×20=100 m\text{Distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 20 = 100 \text{ m}

Worked Example: Multi-Stage v-t Graph

Problem

Question: A v-t graph shows:

  • 0–5 s: velocity increases from 0 to 15 m/s
  • 5–15 s: velocity stays at 15 m/s
  • 15–20 s: velocity decreases from 15 to 0 m/s

Calculate: (a) acceleration in first stage, (b) deceleration in third stage, (c) total distance.

Solution

(a) a=(15−0)/5=3a = (15-0)/5 = 3 m/s² (b) a=(0−15)/5=−3a = (0-15)/5 = -3 m/s² → deceleration = 3 m/s² (c) Distance = Triangle + Rectangle + Triangle =12(5)(15)+(10)(15)+12(5)(15)=37.5+150+37.5=225= \frac{1}{2}(5)(15) + (10)(15) + \frac{1}{2}(5)(15) = 37.5 + 150 + 37.5 = 225 m

Worked Example: Using SUVAT (Higher

Problem

Question: A car accelerates uniformly from rest at 2.5 m/s² for 8 seconds. Calculate the distance travelled.

Solution

u=0u = 0, a=2.5a = 2.5 m/s², t=8t = 8 s, s=?s = ?

s=ut+12at2=0+12(2.5)(82)=12(2.5)(64)=80 ms = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2.5)(8^2) = \frac{1}{2}(2.5)(64) = 80 \text{ m}

Worked Example: Finding Final Velocity (Higher

Problem

Question: A stone is dropped from rest and falls 20 m. Calculate its final velocity. (g=9.8g = 9.8 m/s²)

Solution

u=0u = 0, a=9.8a = 9.8 m/s², s=20s = 20 m, v=?v = ?

v2=u2+2as=0+2(9.8)(20)=392v^2 = u^2 + 2as = 0 + 2(9.8)(20) = 392 v=392=19.8 m/sv = \sqrt{392} = 19.8 \text{ m/s}

7. Practice Questions

    1. A train accelerates from 5 m/s to 35 m/s in 20 seconds. Calculate the acceleration. (2 marks)
    1. A ball decelerates from 12 m/s to rest in 4 seconds. Calculate the deceleration. (2 marks)
    1. A v-t graph shows constant velocity of 8 m/s for 30 seconds. Calculate the distance. (1 mark)
    1. A car accelerates uniformly from rest to 24 m/s in 12 seconds, then travels at 24 m/s for 20 seconds. Calculate the total distance. (4 marks)
    1. (Higher) A cyclist accelerates from 4 m/s at 0.8 m/s² for 10 seconds. Calculate (a) the final velocity, (b) the distance. (4 marks)

    Answers

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Frequently Asked Questions

Is deceleration negative acceleration?

Yes. Deceleration means the object is slowing down. Mathematically, the acceleration is negative (opposite to the direction of motion).

Can velocity be negative?

Yes — negative velocity means the object is moving in the opposite direction to the positive direction you've defined.

How do I handle area under a v-t graph when velocity is negative?

Area below the time axis represents distance travelled in the opposite direction. For total distance, take the absolute values.

Summary

  • Velocity = speed + direction (vector)
  • Acceleration = rate of change of velocity: a=Δv/Δta = \Delta v / \Delta t
  • On v-t graphs: gradient = acceleration, area = distance
  • Positive gradient = accelerating; negative gradient = decelerating
  • Horizontal line = constant velocity; steeper = more acceleration
  • SUVAT equations apply for uniform acceleration (Higher tier)
Tags:
#gcse#physics#forces#velocity#acceleration#v-t-graphs

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