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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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In a Distance-Time graph, the gradient (slope) represents the speed of the object. A straight diagonal line indicates constant speed, while a horizontal line shows the object is stationary. A curve indicates changing speed (acceleration or deceleration).

A distance-time graph showing a straight line starting from the origin, representing constant speed.
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The gradient of a Velocity-Time graph represents the acceleration of the object. If the gradient is positive, the object is accelerating; if negative, it is decelerating. A horizontal line indicates constant velocity (zero acceleration).

A velocity-time graph showing a linear increase in velocity over time, representing constant acceleration.
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The area under a Velocity-Time graph is equal to the total displacement (or distance in a single direction) traveled by the object. For complex shapes, the area can be split into rectangles and triangles to simplify calculations.

A velocity-time graph with shaded regions (triangle and rectangle) showing how to calculate displacement.
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Average speed for a whole journey can be calculated by dividing the total distance by the total time taken, regardless of individual speed changes represented on the graph.

📐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}.

Problem 4:

A cyclist travels along a straight road. Their motion is captured in the provided distance-time graph. Determine the speed of the cyclist between t=2 st = 2 \text{ s} and t=6 st = 6 \text{ s}.

Distance-time graph with points A at (2,20) and B at (6,60).

Solution:

Speed=Change in DistanceChange in Time\text{Speed} = \frac{\text{Change in Distance}}{\text{Change in Time}} Speed=60−206−2\text{Speed} = \frac{60 - 20}{6 - 2} Speed=404\text{Speed} = \frac{40}{4} Speed=10 m/s\text{Speed} = 10 \text{ m/s}

Explanation:

The speed is the gradient of the distance-time graph. By selecting the coordinates (2,20)(2, 20) and (6,60)(6, 60), we calculate the change in distance over the change in time.

Problem 5:

An object accelerates from rest to 12 m/s12 \text{ m/s} in 4 s4 \text{ s}, then maintains this velocity for 4 s4 \text{ s}. Calculate the total distance traveled during these 8 s8 \text{ s} using the velocity-time graph.

Velocity-time graph showing a triangle from 0 to 4 seconds and a rectangle from 4 to 8 seconds.

Solution:

Distance=Area of Triangle+Area of Rectangle\text{Distance} = \text{Area of Triangle} + \text{Area of Rectangle} Area 1=12×base×height=12×4×12=24 m\text{Area 1} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 12 = 24 \text{ m} Area 2=width×height=(8−4)×12=4×12=48 m\text{Area 2} = \text{width} \times \text{height} = (8 - 4) \times 12 = 4 \times 12 = 48 \text{ m} Total Distance=24+48=72 m\text{Total Distance} = 24 + 48 = 72 \text{ m}

Explanation:

The distance traveled is represented by the total area under the velocity-time graph. We split the shape into a triangle (for the acceleration phase) and a rectangle (for the constant velocity phase).