Acceleration and Velocity-Time Graphs Review

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Flashcards covering physics concepts of acceleration, velocity-time graphs, and distance calculations based on lecture exercises.

Last updated 8:09 AM on 6/5/26
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15 Terms

1
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What is a standard unit used to measure acceleration?

m/s2m/s^2

2
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On a speed-time graph, what does a horizontal line indicate about an object's motion?

The object is moving with constant speed.

3
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On a speed-time graph, what physical quantity is represented by the area under the line?

Distance travelled

4
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According to Fig. 2.1, what is the speed of cyclist B?

20m/s20\,m/s

5
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Based on the first 20s of the journey for cyclist B moving at a constant 20m/s20\,m/s, what is the total distance travelled?

distance travelled=20m/s×20s=400m\text{distance travelled} = 20\,m/s \times 20\,s = 400\,m

6
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On a distance-time graph, how is an object moving with constant speed represented?

A straight diagonal line with a constant gradient (labeled A in the exercise).

7
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On a speed-time graph, what type of line represents an object moving with constant acceleration?

A straight diagonal line (labeled S in the exercise).

8
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In Fig. 1.1, which part of the graph shows a car at rest?

Part E

9
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How can you determine from a speed-time graph (like Fig. 1.1) which car has a greater acceleration?

By comparing the gradient (slope) of the lines; a steeper gradient indicates a greater acceleration.

10
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In Fig. 2.1 for the ship's motion, what does section B represent?

Constant speed

11
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What is the initial speed and uniform deceleration rate of the object in Section 3, Task 3, 1(b)?

Initial speed of 50m/s50\,m/s and deceleration of 0.35m/s20.35\,m/s^2.

12
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According to Fig. 1.1 of the roller skater's journey, what letter represents the 'start of motion' characterized by acceleration?

W

13
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If a runner moves at a constant speed of 10m/s10\,m/s for 5.0s5.0\,s, what is the distance travelled?

distance=10m/s×5.0s=50m\text{distance} = 10\,m/s \times 5.0\,s = 50\,m

14
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If a car's speed increases from 0m/s0\,m/s to 25m/s25\,m/s at a uniform rate over 5.0s5.0\,s, what is the distance travelled?

distance=12×base×height=12×5.0s×25m/s=62.5m\text{distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5.0\,s \times 25\,m/s = 62.5\,m

15
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On a speed-time graph, how is an object accelerating at a changing rate or 'increasing acceleration' represented?

By a curved line (labeled T in the exercise).