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Where are we now and where might we be going - Motion: Introducing Graphs

Grade 7IB

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

🔑Concepts

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Distance-Time Graphs: These graphs represent how far an object has travelled in a given time. Time is plotted on the xx-axis and Distance is plotted on the yy-axis.

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The gradient (slope) of a Distance-Time graph represents the speed of the object. A steeper slope indicates a higher speed: Speed=Gradient\text{Speed} = \text{Gradient}.

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A horizontal line on a Distance-Time graph indicates that the object is stationary (v=0 m/sv = 0\text{ m/s}), as the distance does not change as time passes.

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A straight diagonal line on a Distance-Time graph indicates constant speed, meaning the object covers equal distances in equal intervals of time.

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A curved line on a Distance-Time graph indicates acceleration or deceleration. If the curve gets steeper, the object is speeding up; if it levels off, the object is slowing down.

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Speed-Time Graphs: These graphs show how the speed of an object changes over time. The gradient represents acceleration, and the area under the line represents the total distance travelled.

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Scalar vs. Vector: Distance is a scalar quantity (magnitude only), while displacement is a vector quantity (magnitude and direction). Similarly, speed is scalar and velocity is vector.

📐Formulae

Speed(v)=Distance(d)Time(t)\text{Speed} (v) = \frac{\text{Distance} (d)}{\text{Time} (t)}

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

Gradient=RiseRun=y2−y1x2−x1\text{Gradient} = \frac{\text{Rise}}{\text{Run}} = \frac{y_2 - y_1}{x_2 - x_1}

Acceleration(a)=Change in SpeedTime taken=v−ut\text{Acceleration} (a) = \frac{\text{Change in Speed}}{\text{Time taken}} = \frac{v - u}{t}

💡Examples

Problem 1:

A car travels from point AA to point BB. The distance-time graph shows a straight line starting at (0,0)(0, 0) and passing through (5 s,100 m)(5\text{ s}, 100\text{ m}). Calculate the speed of the car.

Solution:

Speed=Change in DistanceChange in Time\text{Speed} = \frac{\text{Change in Distance}}{\text{Change in Time}} Speed=100 m−0 m5 s−0 s\text{Speed} = \frac{100\text{ m} - 0\text{ m}}{5\text{ s} - 0\text{ s}} Speed=1005=20 m/s\text{Speed} = \frac{100}{5} = 20\text{ m/s}

Explanation:

Since the graph is a straight line, the speed is constant. We find the gradient by dividing the total distance (rise) by the total time (run).

Problem 2:

An athlete runs 400 m400\text{ m} in 50 seconds50\text{ seconds}. What is their average speed in m/s\text{m/s}?

Solution:

v=dtv = \frac{d}{t} v=400 m50 sv = \frac{400\text{ m}}{50\text{ s}} v=8 m/sv = 8\text{ m/s}

Explanation:

To find the average speed, we divide the total distance covered by the total time taken for the journey.

Problem 3:

Interpret the motion of an object if its Distance-Time graph is a horizontal line at y=10 my = 10\text{ m} for 5 seconds5\text{ seconds}.

Solution:

Speed=10 m−10 m5 s−0 s=0 m/s\text{Speed} = \frac{10\text{ m} - 10\text{ m}}{5\text{ s} - 0\text{ s}} = 0\text{ m/s}

Explanation:

Because the distance does not change over the 55-second interval, the object is not moving. It is stationary at a position 10 meters10\text{ meters} away from the starting point.