Measuring Acceleration from Velocity-Time Data

When an object moves with changing velocity, we can collect data by measuring the velocity at different time intervals. The acceleration is then found using:

$a = \frac{v - u}{t}$

Where $v$ is final velocity, $u$ is initial velocity, and $t$ is time taken.

Worked Example

A motorcycle on a straight road in Dar es Salaam starts from rest and reaches a speed of 20 m/s after 5 seconds.

Given: $u = 0$ m/s, $v = 20$ m/s, $t = 5$ s

Find: Acceleration

Solution:

$a = \frac{20 - 0}{5} = \frac{20}{5} = 4 \, \text{m/s}^2$

The motorcycle's acceleration is 4 m/s², meaning its velocity increases by 4 m/s every second.

Using the Equations of Motion

When collecting motion data, we can use the three equations of motion to analyse and predict motion:

1. First equation: $v = u + at$ — relates velocity, acceleration, and time
2. Second equation: $s = ut + \frac{1}{2}at^2$ — relates displacement, initial velocity, acceleration, and time
3. Third equation: $v^2 = u^2 + 2as$ — relates velocity, acceleration, and displacement

Worked Example Using All Three Equations

A taxi in Arusha accelerates uniformly from 10 m/s to 30 m/s while covering a distance of 200 m. Find:

• (a) The acceleration
• (b) The time taken
• (c) The final velocity using the third equation (check)

Solution:

Given: $u = 10$ m/s, $v = 30$ m/s, $s = 200$ m

(a) Find acceleration using $v^2 = u^2 + 2as$:

$30^2 = 10^2 + 2 \times a \times 200$

$900 = 100 + 400a$

$800 = 400a$

$a = 2 \, \text{m/s}^2$

(b) Find time using $v = u + at$:

$30 = 10 + 2t$

$20 = 2t$

$t = 10 \, \text{s}$

(c) Verify using $s = ut + \frac{1}{2}at^2$:

$s = 10 \times 10 + \frac{1}{2} \times 2 \times 10^2$

$s = 100 + 100 = 200 \, \text{m}$

The calculation matches the given distance, confirming the data is consistent.

Collecting Friction Data

In a friction experiment, you would collect data by:

1. Measuring the weight (normal force) of a block using a spring balance
2. Pulling the block horizontally with increasing force until it starts moving
3. Recording the force at which motion begins (limiting friction)

Analysing Friction Using the First Law of Friction

The frictional force is directly proportional to the normal force:

$F_f = \mu N$

Where $\mu$ is the coefficient of friction.

Worked Example

A wooden crate with mass 20 kg is placed on a concrete floor in a warehouse in Mwanza. The coefficient of friction is 0.4. Calculate the frictional force.

Solution:

First, find the normal force (weight):

$N = mg = 20 \times 9.8 = 196 \, \text{N}$

Then calculate friction:

$F_f = \mu N = 0.4 \times 196 = 78.4 \, \text{N}$

This means you must pull with more than 78.4 N to start moving the crate.

Analysing Friction Data: Checking the Second and Third Laws

When collecting friction data, you can verify the other laws:

• Second law: Change the orientation of the block (flat or on edge). Even with different contact areas, if $N$ and surface type remain the same, $F_f$ stays constant.
• Third law: Pull the block at different speeds. The kinetic friction force remains approximately the same regardless of velocity.

When collecting motion data experimentally, plotting velocity against time helps us analyse motion visually.

How to Analyse the Graph

• Slope of the line = acceleration (or deceleration if negative)
• Area under the graph = displacement (distance travelled)

Worked Example

A bus in Dodoma starts from rest, accelerates for 30 seconds to reach 15 m/s, then travels at constant speed for 60 seconds, and finally brakes to rest in 15 seconds.

From the data:

• Acceleration phase: slope = $15/30 = 0.5$ m/s²
• Constant speed: slope = 0 (horizontal line)
• Braking phase: slope = $-15/15 = -1$ m/s² (deceleration)

Find total distance (area under graph):

• Triangle 1: $\frac{1}{2} \times 30 \times 15 = 225$ m
• Rectangle: $60 \times 15 = 900$ m
• Triangle 2: $\frac{1}{2} \times 15 \times 15 = 112.5$ m

Total distance: $225 + 900 + 112.5 = 1237.5$ m

When collecting data for objects in free fall (for example, dropping a stone from different heights), we apply the equations with acceleration $g = 9.8$ m/s².

Worked Example

A stone is dropped from a tower in Mbeya and hits the ground after 3 seconds. Find:

• (a) The height of the tower
• (b) The velocity just before impact

Solution:

Given: $u = 0$, $t = 3$ s, $g = 9.8$ m/s²

(a) Height using $h = \frac{1}{2}gt^2$:

$h = \frac{1}{2} \times 9.8 \times 3^2 = 4.9 \times 9 = 44.1 \, \text{m}$

(b) Final velocity using $v = gt$:

$v = 9.8 \times 3 = 29.4 \, \text{m/s}$

1. Identify what is given — list all known quantities from the data
2. Identify what is required — determine which physical parameter you need to find
3. Select the appropriate formula — choose the equation that relates the given quantities to the required one
4. Substitute carefully — put in the values with correct units
5. Calculate and check — work out the answer and verify it makes physical sense

In Tanzania, mechanics and engineers use these data analysis skills when testing vehicle brakes or checking road safety. For example, when a bus company in Dar es Salaam tests braking performance, they measure the stopping distance at different speeds and use the equations of motion to calculate deceleration, ensuring the brakes meet safety standards. Similarly, farmers in rural areas use understanding of friction to select appropriate tyres for tractors working on muddy or dry soil.