G12 T2 Class Work 1
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From the graph provided:When is the bus speeding up?
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From 0 to A and from B to C
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From A to B and from D to E
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From 0 to A and from D to E
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From B to C only
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G12 T2 Class work 1 assesses understanding of bus motion through graph analysis. Students identify intervals of acceleration, calculate specific accelerations, and determine highest acceleration phases, enhancing graph interpretation skills in Physics.
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- 2.
From the graph provided:When is the acceleration 0?
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From 0 to A and from B to C
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From A to B, from C to D and from E to F
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The acceleration is never zero
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From 0 to A, from C to D and from E to F
Correct Answer
A. From A to B, from C to D and from E to F -
- 3.
From the graph provided:When is the bus slowing down?
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From 0 to A and from B to C
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From A to B and from D to E
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From 0 to A and from D to E
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From B to C only
Correct Answer
A. From B to C onlyExplanation
The graph provided shows the speed of the bus over time. From the graph, it can be observed that the bus is slowing down only from point B to point C. This is indicated by the decreasing slope of the graph during that interval. Therefore, the correct answer is "From B to C only".Rate this question:
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- 4.
From the graph provided:Calculate the acceleration between D and E.
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1 m/s2
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0.24 m/s2
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21 m/s2
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5 m/s2
Correct Answer
A. 1 m/s2Explanation
The acceleration between two points on a graph can be calculated by finding the change in velocity and dividing it by the time taken. In this case, the change in velocity between points D and E is 1 m/s and the time taken is 1 second. Therefore, the acceleration is 1 m/s^2.Rate this question:
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- 5.
From the graph provided:State in which interval the acceleration is the highest.
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From 0 to A
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From A to B
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From B to C
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From C to D
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From D to E
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From E to F
Correct Answer
A. From B to CExplanation
The correct answer is From B to C. This can be determined by observing the graph and identifying the steepest slope. The steepest slope represents the highest acceleration. In this case, the steepest slope occurs between points B and C, indicating that the acceleration is the highest in this interval.Rate this question:
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- 6.
Calculate the acceleration for the interval 10 to 15 s.
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0.8 m/s2
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- 2 m/s2
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- 0.8 m/s2
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2 m/s2
Correct Answer
A. - 2 m/s2Explanation
The negative sign in front of the acceleration value indicates that the object is experiencing deceleration or slowing down during the interval from 10 to 15 seconds. The magnitude of the acceleration is 2 m/s2, meaning that the object is slowing down at a rate of 2 meters per second squared.Rate this question:
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- 7.
Calculate the acceleration for the interval 5 to 10 s.
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2.4 m/s2
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1.2 m/s2
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- 1.2 m/s2
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-2.4 m/s2
Correct Answer
A. 1.2 m/s2Explanation
The acceleration for the interval 5 to 10 seconds is calculated to be 1.2 m/s2. This means that the velocity of the object is increasing by 1.2 meters per second every second during this time interval.Rate this question:
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- 8.
The graph represents the movement of a body:How far has the body moved after 5 s?
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2 m
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10 m
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25 m
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50 m
Correct Answer
A. 25 mExplanation
The graph represents the distance traveled by the body over time. After 5 seconds, the graph shows that the body has moved a distance of 25 meters.Rate this question:
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- 9.
Two distance/time graphs and two speed/time graphs are shown.Which graph represents an object that is at rest?
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A
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B
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C
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D
Correct Answer
A. BExplanation
Graph B represents an object at rest because the distance/time graph is a straight horizontal line, indicating that the object is not changing its position over time. Additionally, the speed/time graph for graph B is a flat line at zero, indicating that the object has zero speed and is not moving.Rate this question:
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- 10.
A snail moves along a ruler. It takes 20 s to move from Q to R.What is its average speed from Q to R?
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12/20 cm/s
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(12-2)/20 cm/s
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20/12 cm/s
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20/(12-2) cm/s
Correct Answer
A. (12-2)/20 cm/sExplanation
The snail takes 20 seconds to move from Q to R. The average speed is calculated by dividing the distance traveled by the time taken. Since the distance is not given, we cannot determine the average speed accurately. However, the answer (12-2)/20 cm/s suggests that the snail traveled a distance of 10 cm (12-2) in 20 seconds, resulting in an average speed of 0.5 cm/s.Rate this question:
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Mar 21, 2023Quiz Edited by
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Mar 15, 2015Quiz Created by
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