The graph below models the velocity of a small train as it moves on a straight track for 20 seconds. The front of the train is at the point A when t = 0. The mass of the train is 800kg. The front of the train is at the point A when t = 0. The graph shows the velocity (v) in meters per second (m/s) as a function of time (t) in seconds. 14 (a) Find the total distance travelled in the 20 seconds. 14 (b) Find the distance of the front of the train from the point A at the end of the 20 seconds. 14 (c) Find the maximum magnitude of the resultant force acting on the train. 14 (d) Explain why, in reality, the graph may not be an accurate model of the motion of the train.

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The graph below models the velocity of a small train as it moves on a straight track for 20 seconds - AQA - A-Level Maths Pure - Question 14 - 2017 - Paper 2

Question 14

The graph below models the velocity of a small train as it moves on a straight track for 20 seconds. The front of the train is at the point A when t = 0. The mass ... show full transcript

Worked Solution & Example Answer:The graph below models the velocity of a small train as it moves on a straight track for 20 seconds - AQA - A-Level Maths Pure - Question 14 - 2017 - Paper 2

Step 1

Find the total distance travelled in the 20 seconds.

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Answer

To calculate the total distance travelled, we need to find the area under the velocity-time graph. The graph consists of several segments:

From t = 0 to t = 6 seconds, the velocity is constant at 8 m/s. Therefore, the distance is: $s_1 = v imes t = 8 imes 6 = 48 ext{ m}$
From t = 6 to t = 10 seconds, the velocity decreases linearly to 0 m/s. The area can be calculated as a triangle: $s_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 8 = 16 ext{ m}$
From t = 10 to t = 20 seconds, the velocity is constant at 0 m/s, which contributes no distance: $s_3 = 0 ext{ m}$

Adding these distances together gives: $ext{Total Distance} = s_1 + s_2 + s_3 = 48 + 16 + 0 = 64 ext{ m}$

Step 2

Find the distance of the front of the train from the point A at the end of the 20 seconds.

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Answer

From part (a), we found that the total distance travelled in 20 seconds is 64 m. Therefore, the front of the train is 64 m from point A at the end of the 20 seconds.

Step 3

Find the maximum magnitude of the resultant force acting on the train.

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Answer

To determine the maximum resultant force, we first need to find the maximum acceleration, which occurs when the train's velocity decreases. The maximum acceleration can be calculated using the change in velocity over time:

The maximum change in velocity is from 8 m/s to 0 m/s over the time interval from 6 to 10 seconds: $a_{max} = \frac{\Delta v}{\Delta t} = \frac{0 - 8}{10 - 6} = \frac{-8}{4} = -2 ext{ m/s}^2$

Now, using Newton's second law, the force is calculated as: $F = m \times a = 800 imes (-2) = -1600 ext{ N}$

Thus, the maximum magnitude of the resultant force is 1600 N.

Step 4

Explain why, in reality, the graph may not be an accurate model of the motion of the train.

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Answer

The graph presents abrupt changes in velocity and straight lines, which are unrealistic in real train motion. In practice, trains undergo gradual acceleration and deceleration due to factors such as friction, inertia, and mechanical constraints. Such factors would create curves in the graph that represent a more realistic acceleration profile, rather than the sudden shifts illustrated.

The graph below models the velocity of a small train as it moves on a straight track for 20 seconds - AQA - A-Level Maths Pure - Question 14 - 2017 - Paper 2

Worked Solution & Example Answer:The graph below models the velocity of a small train as it moves on a straight track for 20 seconds - AQA - A-Level Maths Pure - Question 14 - 2017 - Paper 2

Find the total distance travelled in the 20 seconds.

Find the distance of the front of the train from the point A at the end of the 20 seconds.

Find the maximum magnitude of the resultant force acting on the train.

Explain why, in reality, the graph may not be an accurate model of the motion of the train.

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