Speed Time Graphs 5 worksheet featuring four physics problems on motion, including drawing speed-time graphs and calculating distance traveled.

Problem: Speed-Time Graphs 5

We will solve each part of the problem step by step.

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#### Question 1:
A car accelerates from rest to a speed of 10 m/s in 4 seconds. It then decelerates to rest in 2 seconds.
1. Draw the speed-time graph for the car.
2. Work out the total distance travelled.

##### Solution:
1. Speed-Time Graph:
- The car starts from rest, so the initial speed is 0 m/s.
- It accelerates uniformly to 10 m/s in 4 seconds.
- Then it decelerates uniformly to rest in 2 seconds.
- The graph will have two segments:
- A straight line from (0, 0) to (4, 10) representing acceleration.
- A straight line from (4, 10) to (6, 0) representing deceleration.

2. Total Distance Travelled:
- The distance travelled is the area under the speed-time graph.
- The graph forms a trapezium.
- Area of the trapezium:
\[
\text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times (\text{Height})
\]
- Parallel sides: 0 m/s and 10 m/s.
- Height: 4 seconds (acceleration phase).
- Area of the first triangle (acceleration):
\[
\text{Area}_1 = \frac{1}{2} \times 10 \times 4 = 20 \text{ m}
\]
- Area of the second triangle (deceleration):
\[
\text{Area}_2 = \frac{1}{2} \times 10 \times 2 = 10 \text{ m}
\]
- Total distance:
\[
\text{Total Distance} = \text{Area}_1 + \text{Area}_2 = 20 + 10 = 30 \text{ m}
\]

Answer for Question 1:
\[
\boxed{30 \text{ m}}
\]

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#### Question 2:
A cyclist started from rest and accelerated at 1.5 m/s² for 5 seconds and then travelled at a constant speed.
1. What was the speed of the cyclist after 5 seconds?
2. Draw the speed-time graph for the cyclist.

##### Solution:
1. Speed of the Cyclist After 5 Seconds:
- Initial speed, \( u = 0 \) m/s.
- Acceleration, \( a = 1.5 \) m/s².
- Time, \( t = 5 \) seconds.
- Final speed, \( v \), can be calculated using:
\[
v = u + at = 0 + 1.5 \times 5 = 7.5 \text{ m/s}
\]

2. Speed-Time Graph:
- The graph will have two segments:
- A straight line from (0, 0) to (5, 7.5) representing acceleration.
- A horizontal line from (5, 7.5) onwards representing constant speed.

Answer for Question 2:
\[
\boxed{7.5 \text{ m/s}}
\]

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#### Question 3:
A car accelerates uniformly from rest to a speed of 24 m/s after accelerating at 2 m/s². It travels at 24 m/s for 30 seconds before slowing down steadily to rest. The whole journey takes 60 seconds.
1. Draw the speed-time graph for the car.
2. Work out the rate of deceleration.

##### Solution:
1. Speed-Time Graph:
- The car starts from rest and accelerates uniformly to 24 m/s.
- Acceleration, \( a = 2 \) m/s².
- Final speed, \( v = 24 \) m/s.
- Time to reach 24 m/s:
\[
v = u + at \implies 24 = 0 + 2t \implies t = 12 \text{ seconds}
\]
- The car then travels at a constant speed of 24 m/s for 30 seconds.
- Finally, it decelerates to rest in the remaining time:
\[
\text{Total time} = 60 \text{ seconds}, \quad \text{Time for deceleration} = 60 - 12 - 30 = 18 \text{ seconds}
\]
- The graph will have three segments:
- A straight line from (0, 0) to (12, 24) representing acceleration.
- A horizontal line from (12, 24) to (42, 24) representing constant speed.
- A straight line from (42, 24) to (60, 0) representing deceleration.

2. Rate of Deceleration:
- Initial speed during deceleration, \( u = 24 \) m/s.
- Final speed, \( v = 0 \) m/s.
- Time, \( t = 18 \) seconds.
- Deceleration, \( a \), is given by:
\[
v = u + at \implies 0 = 24 + a \times 18 \implies a = -\frac{24}{18} = -\frac{4}{3} \text{ m/s}^2
\]

Answer for Question 3:
\[
\boxed{-\frac{4}{3} \text{ m/s}^2}
\]

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#### Question 4:
A motorcyclist accelerates from rest at 5 m/s² for 8 seconds, then travels at a constant speed for 10 seconds. It then decelerates at 3 m/s² for 4 seconds and continues at constant speed.
1. Draw a speed-time graph for the motorcyclist.
2. Find the total distance travelled when the motorcyclist decelerated.

##### Solution:
1. Speed-Time Graph:
- The motorcyclist starts from rest and accelerates uniformly at 5 m/s² for 8 seconds.
- Final speed after acceleration:
\[
v = u + at = 0 + 5 \times 8 = 40 \text{ m/s}
\]
- The motorcyclist then travels at a constant speed of 40 m/s for 10 seconds.
- Next, the motorcyclist decelerates at 3 m/s² for 4 seconds.
- Final speed after deceleration:
\[
v = u + at = 40 + (-3) \times 4 = 40 - 12 = 28 \text{ m/s}
\]
- The graph will have four segments:
- A straight line from (0, 0) to (8, 40) representing acceleration.
- A horizontal line from (8, 40) to (18, 40) representing constant speed.
- A straight line from (18, 40) to (22, 28) representing deceleration.
- A horizontal line from (22, 28) onwards representing constant speed.

2. Total Distance Travelled During Deceleration:
- The distance travelled during deceleration is the area under the speed-time graph for the deceleration phase.
- This forms a trapezium with:
- Parallel sides: 40 m/s and 28 m/s.
- Height: 4 seconds.
- Area of the trapezium:
\[
\text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times (\text{Height})
\]
\[
\text{Area} = \frac{1}{2} \times (40 + 28) \times 4 = \frac{1}{2} \times 68 \times 4 = 136 \text{ m}
\]

Answer for Question 4:
\[
\boxed{136 \text{ m}}
\]

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Final Answers:

1. \(\boxed{30 \text{ m}}\)
2. \(\boxed{7.5 \text{ m/s}}\)
3. \(\boxed{-\frac{4}{3} \text{ m/s}^2}\)
4. \(\boxed{136 \text{ m}}\)