Speed Velocity and Acceleration
Kinematics is the study of how objects move, describing their path, speed, and acceleration without considering the forces involved. It is a foundational topic for solving almost all mechanics problems in the ESAT.
Part of the ESAT Physics syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Vector quantities (e.g., displacement, velocity) have both magnitude and direction, while scalar quantities (e.g., distance, speed) have only magnitude.
- The direction of a vector is crucial; a negative sign typically indicates the opposite direction to the one defined as positive.
- On a velocity-time graph, the gradient represents acceleration, and the area under the graph represents displacement.
- On a displacement-time graph, the gradient represents velocity.
- The equation v² − u² = 2as is only valid for motion with constant acceleration.
Diagram
› Why does this happen?
Why is the gradient on a graph so important?
The 'gradient' of a line is simply a measure of its steepness, calculated as 'rise over run'. The 'rise' is the change on the vertical (y) axis, and the 'run' is the change on the horizontal (x) axis. In physics graphs, these axes represent physical quantities. For a displacement-time graph, the gradient is (change in displacement) / (change in time), which is the definition of velocity. For a velocity-time graph, the gradient is (change in velocity) / (change in time), which is the definition of acceleration. So, calculating the gradient is a visual way of applying the definition.
Why is the area under a velocity-time graph displacement?
For an object moving at a constant velocity 'v' for a time 't', its displacement is simply s = v × t. On a velocity-time graph, this motion is a horizontal line. The area under this line is a rectangle with height 'v' and width 't', so its area (v × t) is the displacement. If acceleration is constant, the graph is a sloped line, and the area under it is a trapezium. The kinematics equation s = (u+v)/2 × t directly matches the formula for a trapezium's area. Here, the initial and final velocities ('u' and 'v') are the parallel sides, and the time ('t') is the height between them.
Formulae
average speed = total distance / total time For any journey, especially those with varying speeds. This is a very common calculation.
velocity = change in displacement / time For calculating average velocity or when velocity is constant. Remember displacement is a vector.
acceleration = change in velocity / time For calculating average acceleration or when acceleration is constant.
v2 - u2 = 2as For motion under constant acceleration when time (t) is not known or not required. Here, u is initial velocity, v is final velocity, a is acceleration, and s is displacement.
Definitions
- Displacement (s)
- A vector quantity representing the shortest straight-line distance and direction from an object's starting point to its finishing point.
- Distance
- A scalar quantity representing the total length of the path travelled by an object, regardless of direction.
- Velocity (v)
- A vector quantity; the rate of change of displacement. It specifies both speed and direction.
- Speed
- A scalar quantity; the rate of change of distance. It describes how fast an object is moving.
- Acceleration (a)
- A vector quantity; the rate of change of velocity. A negative acceleration (deceleration) means the velocity is decreasing or increasing in the negative direction.
Worked example
A particle moves along a straight line. It starts from rest and accelerates uniformly for 6 seconds, reaching a velocity of 12 m/s. It then immediately decelerates uniformly, coming to rest after travelling a further 18 m. What is the particle's average speed for the entire journey?
- 1
Step 1:
Analyse the first stage (acceleration).
Find the distance travelled (s1).
We have u=0, v=12 m/s, t=6 sThe acceleration is a = (12-0)/6 = 2 m/s².
Now use v² - u² = 2as to find the distance:
12² - 0² = 2 × 2 × s1, which gives 144 = 4 × s1, so s1 = 36 m - 2
Step 2:
Analyse the second stage (deceleration).
The distance (s2) is given as 18 m.
To find the time taken (t2), we can first find the deceleration.
u=12 m/s, v=0, s=18 mUsing v² - u² = 2as:
0² - 12² = 2 × a × 18, so -144 = 36a, giving a = -4 m/s²Now find the time:
a = (v-u)/t → -4 = (0-12)/t2, so t2 = 12/4 = 3 s - 3
Step 3:
Calculate total distance and total time for the entire journey.
Total distance = s1 + s2 = 36 m + 18 m = 54 m.
Total time = t1 + t2 = 6 s + 3 s = 9 s - 4
Step 4:
Calculate the average speed.
Average speed = total distance / total time = 54 m / 9 s.
Answer: 6 m/s
Common mistakes
- ×Confusing distance with displacement, or speed with velocity. For a journey that involves a change in direction, total distance will be greater than the magnitude of the final displacement.
- ×Making sign errors with vectors. Always define a positive direction at the start. If an object moves in the opposite direction, its velocity and displacement are negative.
- ×Incorrectly reading graph axes. Always check if a graph shows distance-time, velocity-time, etc. The meaning of the gradient and area depends entirely on this.
- ×Applying v² - u² = 2as when acceleration is not constant. This formula is only for uniform acceleration.
- ×Forgetting to convert units, particularly minutes to seconds or km to metres, leading to an answer that is off by a factor of 60 or 1000.
No-calculator tips
- ✓When calculating area under a velocity-time graph, split complex shapes into simple rectangles and triangles. The area of a trapezium is 0.5 × (a+b) × h, which can be faster than splitting.
- ✓In calculations like `v² - u² = 2as`, rearrange the formula to isolate the unknown before substituting numbers. This can simplify the arithmetic e.g., `a = (v² - u²) / (2s)`.
- ✓Look for opportunities to cancel numbers in fractions before multiplying. For example, in `(144*5)/12`, notice that 144 is 12*12, so the expression simplifies to `12*5 = 60`.