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Physics: Motion and Mechanics - Measurement, Experimental Uncertainty, Scalars, and Vectors

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Measurement in Science: Standard International (SI) units are used globally to ensure consistency. The base units include the meter (mm) for length, kilogram (kgkg) for mass, and second (ss) for time.

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Accuracy vs. Precision: Accuracy refers to how close a measurement is to the true or accepted value, while precision refers to the consistency or reproducibility of repeated measurements.

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Experimental Uncertainty: Every measurement has an associated uncertainty. Random errors result from unpredictable fluctuations (e.g., reaction time), while Systematic errors result from equipment flaws (e.g., zero error on a scale).

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Scalars: Physical quantities that have only magnitude (size) and no direction. Examples include distance (dd), speed (vv), mass (mm), and time (tt).

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Vectors: Physical quantities that have both magnitude and direction. Examples include displacement (ss), velocity (vv), acceleration (aa), and force (FF).

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Distance vs. Displacement: Distance is the total path length traveled (scalar), whereas displacement is the straight-line change in position from the start to the end point (vector).

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Speed vs. Velocity: Speed is the rate of change of distance (v=dtv = \frac{d}{t}), while velocity is the rate of change of displacement (v⃗=ΔsΔt\vec{v} = \frac{\Delta s}{\Delta t}).

📐Formulae

v=dtv = \frac{d}{t}

v⃗=ΔsΔt\vec{v} = \frac{\Delta s}{\Delta t}

Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

Uncertainty=±Range2\text{Uncertainty} = \pm \frac{\text{Range}}{2}

Percentage Uncertainty=Absolute UncertaintyMeasured Value×100%\text{Percentage Uncertainty} = \frac{\text{Absolute Uncertainty}}{\text{Measured Value}} \times 100\%

💡Examples

Problem 1:

A student measures the length of a desk three times and gets: 1.22 m1.22 \text{ m}, 1.24 m1.24 \text{ m}, and 1.23 m1.23 \text{ m}. Calculate the average length and the absolute uncertainty.

Solution:

Average=1.22+1.24+1.233=1.23 m\text{Average} = \frac{1.22 + 1.24 + 1.23}{3} = 1.23 \text{ m} Uncertainty=1.24−1.222=0.01 m\text{Uncertainty} = \frac{1.24 - 1.22}{2} = 0.01 \text{ m} Result=1.23±0.01 m\text{Result} = 1.23 \pm 0.01 \text{ m}

Explanation:

To find the average, sum the measurements and divide by the count. The uncertainty is often estimated as half the range (max value minus min value).

Problem 2:

An athlete runs 400 m400 \text{ m} around a circular track and returns to the starting point in 50 seconds50 \text{ seconds}. Calculate their average speed and average velocity.

Solution:

Speed=DistanceTime=400 m50 s=8 m/s\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{400 \text{ m}}{50 \text{ s}} = 8 \text{ m/s} Velocity=DisplacementTime=0 m50 s=0 m/s\text{Velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{0 \text{ m}}{50 \text{ s}} = 0 \text{ m/s}

Explanation:

Since the athlete returns to the start, the displacement is 00. Speed depends on the total path length (400 m400 \text{ m}), but velocity depends on the change in position.

Problem 3:

A car travels 60 km60 \text{ km} North and then 20 km20 \text{ km} South. Find the total distance and the final displacement.

Solution:

Distance=60 km+20 km=80 km\text{Distance} = 60 \text{ km} + 20 \text{ km} = 80 \text{ km} Displacement=60 km [N]−20 km [N]=40 km [North]\text{Displacement} = 60 \text{ km [N]} - 20 \text{ km [N]} = 40 \text{ km [North]}

Explanation:

Distance is a scalar sum of all paths. Displacement is a vector sum; since South is the opposite of North, we subtract the magnitudes to find the resultant vector.