Here is a speed-time graph for a train journey between two stations - Edexcel - GCSE Maths - Question 22 - 2019 - Paper 2
Question 22
Here is a speed-time graph for a train journey between two stations.
The journey took 100 seconds.
(a) Calculate the time taken by the train to travel half the dist... show full transcript
Worked Solution & Example Answer:Here is a speed-time graph for a train journey between two stations - Edexcel - GCSE Maths - Question 22 - 2019 - Paper 2
Step 1
Calculate the time taken by the train to travel half the distance between the two stations.
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Answer
To find the distance traveled during the journey, we calculate the area under the speed-time graph.
Calculate the area for the first part (0 to 60 seconds):
The speed is constant at 12 m/s for 60 seconds.
Area = Speed × Time = 12 m/s × 60 s = 720 m.
Calculate the area for the second part (60 to 100 seconds):
The speed changes linearly from 12 m/s to 0 m/s.
This forms a triangle.
Base = 40 s (from 60 to 100 seconds), Height = 12 m/s.
Area = (1/2) × Base × Height = 0.5 × 40 s × 12 m/s = 240 m.
Total distance = First part + Second part = 720 m + 240 m = 960 m.
Half the distance = 960 m / 2 = 480 m.
To find the time taken to travel 480 m:
During the first part (60 seconds), the train traveled 720 m. Hence, the entire journey's 480 m is fully accommodated within the first segment.
To find an exact point at which the 480 m occurs, we can calculate from the speed.
Using proportion:
Time for 480 m = (480 m / 12 m/s) = 40 seconds.
Thus, the answer for part (a) is 40 seconds.
Step 2
Compare the acceleration of the train during the first part of its journey with the acceleration of the train during the last part of its journey.
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Answer
In the first part of the journey (0 to 60 seconds), the train moves with a constant speed of 12 m/s, resulting in an acceleration of 0 m/s² since there is no change in speed.
In the last part of the journey (60 to 100 seconds), the train decelerates from 12 m/s to 0 m/s.
The initial speed is 12 m/s, and the final speed is 0 m/s over a duration of 40 seconds.
Thus, the acceleration can be calculated using the formula:
a=tvf−vi=400−12=−0.3 m/s2
Therefore, the acceleration is negative during the last part, indicating deceleration, while the acceleration during the first part is 0 m/s².