Interpreting Speed-Time Graphs Worksheet | PDF Printable Algebra ... - Free Printable

Let's solve each problem step by step, interpreting the speed-time graphs.

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## Problem 1: Train slowing down

Graph: Straight line from (0, 30) to (60, 0)

a) What is the deceleration of the train?

Deceleration = rate of decrease in speed = slope of the graph (negative because it’s slowing down).

Slope = (change in speed) / (change in time)
= (0 m/s - 30 m/s) / (60 s - 0 s)
= (-30) / 60
= -0.5 m/s²

So, deceleration = 0.5 m/s² (magnitude; deceleration is positive when referring to rate of slowing).

✔ Answer: 0.5 m/s²

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b) What is the distance travelled?

In a speed-time graph, distance = area under the curve.

This is a triangle:

Base = 60 s
Height = 30 m/s

Area = (1/2) × base × height
= (1/2) × 60 × 30
= 900 meters

✔ Answer: 900 m

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## Problem 2: Car’s 60-second journey

Graph: Trapezoid shape — rises to 15 m/s at 20 s, stays constant until 50 s, then drops to 0 at 60 s.

a) What is the maximum speed of the car?

Highest point on graph = 15 m/s

✔ Answer: 15 m/s

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b) What is the initial acceleration?

From t=0 to t=20 s: speed goes from 0 to 15 m/s.

Acceleration = Δv / Δt = (15 - 0) / (20 - 0) = 15/20 = 0.75 m/s²

✔ Answer: 0.75 m/s²

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c) What is the final deceleration?

From t=50 to t=60 s: speed drops from 15 to 0 m/s.

Deceleration = |Δv / Δt| = |(0 - 15)/(60 - 50)| = 15/10 = 1.5 m/s²

✔ Answer: 1.5 m/s²

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d) What is the distance travelled?

Area under graph = trapezoid.

Can break into:
- Triangle (0–20 s): (1/2) × 20 × 15 = 150 m
- Rectangle (20–50 s): 30 s × 15 m/s = 450 m
- Triangle (50–60 s): (1/2) × 10 × 15 = 75 m

Total = 150 + 450 + 75 = 675 meters

✔ Answer: 675 m

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e) What is the average speed for the whole journey?

Average speed = total distance / total time
= 675 m / 60 s = 11.25 m/s

✔ Answer: 11.25 m/s

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## Problem 3: Speedboat’s start

Graph: Rises to 22 m/s at t=2s, stays constant until t=6s, then decreases to 10 m/s at t=8s, then constant.

*(Note: From graph, at t=2, speed ≈ 22 m/s; at t=6, still 22 m/s; at t=8, drops to 10 m/s)*

a) Find acceleration between t=0 and t=2.

Δv = 22 - 0 = 22 m/s
Δt = 2 - 0 = 2 s
Acceleration = 22 / 2 = 11 m/s²

✔ Answer: 11 m/s²

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b) Describe motion between t=2 and t=6.

Speed is constant at 22 m/s → constant speed motion (no acceleration)

✔ Answer: The speedboat is moving at a constant speed of 22 m/s.

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c) Distance travelled in first 4 seconds.

Break into:
- 0–2 s: triangle → (1/2) × 2 × 22 = 22 m
- 2–4 s: rectangle → 2 s × 22 m/s = 44 m

Total = 22 + 44 = 66 meters

✔ Answer: 66 m

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## Problem 4: Car and Motorbike

Graph: Two lines — motorbike (green) accelerates to 20 m/s at t=10s, then constant. Car (blue) accelerates to 30 m/s at t=30s, then constant.

They cross at one point — where speeds are equal.

a) When are they travelling at the same speed?

Look for intersection of the two lines.

Motorbike: constant at 20 m/s after t=10s.

Car: accelerating from 0 to 30 m/s over 30 s → slope = 1 m/s² → equation: v = t (since 30/30 = 1)

Set car speed = motorbike speed:

t = 20 → so at t = 20 seconds

✔ Answer: At t = 20 seconds

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b) How far apart are they when they are travelling at the same speed?

We need distance each has travelled at t=20 s.

#### Motorbike:
- 0–10 s: triangle → (1/2) × 10 × 20 = 100 m
- 10–20 s: rectangle → 10 s × 20 m/s = 200 m
→ Total = 100 + 200 = 300 m

#### Car:
Accelerating uniformly: v = t → distance = area under triangle from 0 to 20 s.

Area = (1/2) × base × height = (1/2) × 20 × 20 = 200 m

Difference = 300 - 200 = 100 meters

✔ Answer: 100 meters apart

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## ✔ Final Answers Summary:

1)

a) Deceleration = 0.5 m/s²
b) Distance = 900 m

2)

a) Max speed = 15 m/s
b) Initial acceleration = 0.75 m/s²
c) Final deceleration = 1.5 m/s²
d) Distance = 675 m
e) Average speed = 11.25 m/s

3)

a) Acceleration (0–2s) = 11 m/s²
b) Motion (2–6s) = Constant speed of 22 m/s
c) Distance (first 4s) = 66 m

4)

a) Same speed at t = 20 seconds
b) Distance apart = 100 meters

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Let me know if you’d like visual explanations or formulas broken down further!