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Year 11 Science

Velocity and Acceleration

Understand the difference between speed and velocity, explore instantaneous and average values, and interpret velocity-time graphs to analyse motion.

Speed vs Velocity

Speed is a scalar quantity -- it tells you how fast an object is moving, but not the direction. Velocity is a vector quantity -- it includes both magnitude (speed) and direction. In physics, this distinction is critical.

Average Velocity

vavg = Δx / Δt

The total displacement divided by the total time taken. Displacement is the straight-line distance from start to finish, with direction.

Instantaneous Velocity

v = limΔt→0 Δx/Δt

The velocity at a specific instant in time. On a position-time graph, it equals the gradient of the tangent at that point.

Key distinction: A car travelling around a circular track at a constant speed of 60 km/h has a constantly changing velocity because its direction is always changing.

Acceleration

Acceleration is the rate of change of velocity with respect to time. It is a vector quantity measured in metres per second squared (m s-2). An object accelerates whenever its speed or direction changes.

Equations of Uniform Acceleration (SUVAT)

v = u + at

Final velocity from initial velocity and acceleration

s = ut + ½at²

Displacement from initial velocity, acceleration and time

v² = u² + 2as

Final velocity without needing time

s = ½(u + v)t

Displacement from average of initial and final velocities

Where: s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time

Remember: Negative acceleration (deceleration) means the object is slowing down in its direction of motion. Acceleration due to gravity near Earth's surface is approximately 9.8 m s-2 downward.

Velocity-Time Graphs

Velocity-time (v-t) graphs provide a powerful visual tool for analysing motion. The gradient of the graph gives the acceleration, and the area under the curve gives the displacement.

Reading a Velocity-Time Graph

Positive gradient (line slopes up)

Object is accelerating -- velocity increasing

Zero gradient (horizontal line)

Constant velocity -- no acceleration

Negative gradient (line slopes down)

Object is decelerating -- velocity decreasing

Area under the graph

Equals the displacement of the object

Interpreting Graph Shapes

Straight line from origin: Uniform acceleration from rest -- the velocity increases at a constant rate.

Curved line: Non-uniform (changing) acceleration -- the rate of velocity change itself is changing.

Line below the time axis: The object is moving in the negative direction (opposite to the chosen positive direction).

Key Vocabulary

Displacement

The change in position of an object, measured as a straight-line distance with direction from start to finish. SI unit: metres (m).

Velocity

The rate of change of displacement with respect to time. A vector quantity with SI unit m s-1.

Acceleration

The rate of change of velocity with respect to time. SI unit: m s-2. Can be positive (speeding up) or negative (slowing down).

Uniform Motion

Motion at constant velocity (zero acceleration). On a v-t graph, this appears as a horizontal line.

Worked Examples

1

A car accelerates uniformly from 10 m s-1 to 30 m s-1 in 5 seconds. Calculate the acceleration.

Step 1: Identify known values: u = 10 m s-1, v = 30 m s-1, t = 5 s.

Step 2: Use v = u + at, rearranged to a = (v - u) / t.

Step 3: a = (30 - 10) / 5 = 20 / 5 = 4 m s-2.

Answer: The car accelerates at 4 m s-2.

2

A cyclist travels 400 m north in 50 s, then 300 m south in 30 s. Calculate the average speed and average velocity.

Step 1: Total distance = 400 + 300 = 700 m. Total time = 50 + 30 = 80 s.

Step 2: Average speed = total distance / total time = 700 / 80 = 8.75 m s-1.

Step 3: Total displacement = 400 north - 300 south = 100 m north.

Step 4: Average velocity = displacement / time = 100 / 80 = 1.25 m s-1 north.

3

A car brakes uniformly from 20 m s-1 to rest over a distance of 40 m. Find the deceleration.

Step 1: Known: u = 20 m s-1, v = 0, s = 40 m.

Step 2: Use v² = u² + 2as ⇒ 0 = (20)² + 2a(40).

Step 3: 0 = 400 + 80a ⇒ a = -400/80 = -5 m s-2.

Answer: The deceleration is 5 m s-2 (acceleration = -5 m s-2).

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

What is the SI unit of acceleration?

Question 2

A train accelerates uniformly from rest to 25 m s-1 in 10 s. What is its acceleration?

Question 3

On a velocity-time graph, what does the area under the line represent?

Question 4

Which quantity is a vector?

Question 5

A ball is dropped from rest and falls for 3 s (g = 9.8 m s-2). What is its final velocity?

Key Concepts Summary

Year 10: The Universe Year 11: Projectile Motion