Comprehensive worksheet featuring four speed-time graph problems covering acceleration, deceleration, and distance calculations for GCSE students.

I'll solve each problem step by step, working through the calculations carefully.

## Problem 1: Train slowing down

a) Deceleration of the train:
- Initial speed at t=0: 30 m/s
- Final speed at t=60: 0 m/s
- Time taken: 60 seconds
- Deceleration = (change in speed) / time = (0 - 30) / 60 = -30/60 = -0.5 m/s²
- The deceleration is 0.5 m/s² (we give it as a positive value since "deceleration" already indicates slowing down)

b) Distance travelled:
- Area under the speed-time graph gives distance
- This is a triangle with base = 60 s and height = 30 m/s
- Area = ½ × base × height = ½ × 60 × 30 = 900 m
- Distance = 900 m

## Problem 2: Car's journey (60 seconds)

a) Maximum speed:
Looking at the graph, the highest point is at 15 m/s
- Maximum speed = 15 m/s

b) Initial acceleration:
- From t=0 to t=20, speed goes from 0 to 15 m/s
- Acceleration = (15 - 0) / 20 = 15/20 = 0.75 m/s²

c) Final deceleration:
- From t=50 to t=60, speed goes from 15 to 0 m/s
- Deceleration = (15 - 0) / 10 = 15/10 = 1.5 m/s²

d) Distance travelled:
The area consists of three parts:
1. Triangle (0-20s): ½ × 20 × 15 = 150 m
2. Rectangle (20-50s): 30 × 15 = 450 m
3. Triangle (50-60s): ½ × 10 × 15 = 75 m

Total distance = 150 + 450 + 75 = 675 m

e) Average speed:
- Average speed = Total distance / Total time
- Average speed = 675 / 60 = 11.25 m/s

## Problem 3: Speedboat's journey

a) Acceleration between t=0 and t=2:
- At t=0: speed = 0 m/s
- At t=2: speed = 22 m/s (reading from graph - each small square is 1 unit on y-axis)
- Acceleration = (22 - 0) / 2 = 11 m/s²

b) Motion between t=2 and t=6:
- From t=2 to t=6, the line is horizontal at 22 m/s
- The speedboat is moving at constant speed of 22 m/s

c) Distance in first 4 seconds:
This has two parts:
1. Triangle (0-2s): ½ × 2 × 22 = 22 m
2. Rectangle (2-4s): 2 × 22 = 44 m

Total distance = 22 + 44 = 66 m

## Problem 4: Car and motorbike

a) When are they travelling at the same speed?
Looking at where the lines intersect:
- The car (blue) and motorbike (green) meet at approximately t=20 seconds
- At this point, both are at 20 m/s
- Answer: t = 20 seconds

b) How far apart when travelling at same speed?

Distance for motorbike at t=20:
- Triangle (0-10s): ½ × 10 × 20 = 100 m
- Rectangle (10-20s): 10 × 20 = 200 m
- Total = 300 m

Distance for car at t=20:
- Triangle: ½ × 20 × 20 = 200 m

Distance apart = 300 - 200 = 100 m

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Final Answer:

Problem 1:
a) 0.5 m/s²
b) 900 m

Problem 2:
a) 15 m/s
b) 0.75 m/s²
c) 1.5 m/s²
d) 675 m
e) 11.25 m/s

Problem 3:
a) 11 m/s²
b) Constant speed of 22 m/s
c) 66 m

Problem 4:
a) t = 20 seconds
b) 100 m