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Forces and Motion - Distance-Time and Velocity-Time Graphs

Grade 9IB

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

🔑Concepts

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Distance-Time Graphs: The gradient (slope) of a distance-time graph represents the speed of the object. A steeper gradient indicates a higher speed vv.

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Stationary Objects: On a distance-time graph, a horizontal line (gradient =0= 0) means the object is stationary (v=0 m/sv = 0 \text{ m/s}).

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Velocity-Time Graphs: The gradient of a velocity-time graph represents the acceleration aa of the object. A positive gradient indicates acceleration, while a negative gradient indicates deceleration (retardation).

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Area Under the Graph: The total displacement ss (or distance) traveled is calculated by finding the area under the line in a velocity-time graph.

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Uniform Motion: A straight diagonal line on a distance-time graph indicates constant speed. A straight diagonal line on a velocity-time graph indicates constant (uniform) acceleration.

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Non-Uniform Motion: Curved lines on distance-time graphs indicate that the speed is changing, which means the object is accelerating or decelerating.

📐Formulae

Speed(v)=Δdistance(s)Δtime(t)\text{Speed} (v) = \frac{\Delta \text{distance} (s)}{\Delta \text{time} (t)}

Acceleration(a)=v−ut\text{Acceleration} (a) = \frac{v - u}{t}

Gradient(m)=y2−y1x2−x1\text{Gradient} (m) = \frac{y_2 - y_1}{x_2 - x_1}

Displacement(s)=Area under v-t graph\text{Displacement} (s) = \text{Area under } v\text{-}t \text{ graph}

💡Examples

Problem 1:

A car's motion is plotted on a distance-time graph. It travels 150 m150 \text{ m} in 10 s10 \text{ s} at a constant rate. Calculate its speed.

Solution:

v=150 m10 s=15 m/sv = \frac{150 \text{ m}}{10 \text{ s}} = 15 \text{ m/s}

Explanation:

Since the speed is constant, the gradient of the distance-time graph is ΔyΔx\frac{\Delta y}{\Delta x}, which equals 15 m/s15 \text{ m/s}.

Problem 2:

An athlete starts from rest and reaches a velocity of 8 m/s8 \text{ m/s} in 4 s4 \text{ s} on a velocity-time graph. Find the acceleration.

Solution:

a=8 m/s−0 m/s4 s=2 m/s2a = \frac{8 \text{ m/s} - 0 \text{ m/s}}{4 \text{ s}} = 2 \text{ m/s}^2

Explanation:

Acceleration is the change in velocity divided by time, represented by the gradient of the v−tv-t graph.

Problem 3:

Calculate the total distance traveled for an object that moves at a constant velocity of 10 m/s10 \text{ m/s} for 5 s5 \text{ s}.

Solution:

s=10 m/s×5 s=50 ms = 10 \text{ m/s} \times 5 \text{ s} = 50 \text{ m}

Explanation:

On a velocity-time graph, this motion is represented by a horizontal line. The area of the rectangle formed (base×heightbase \times height) gives the distance: 5 s×10 m/s=50 m5 \text{ s} \times 10 \text{ m/s} = 50 \text{ m}.