Motion Graphs & Kinematic Equations — Drills · Grade 12 Science
Chapter: Motion & KinematicsWhat these drills covers
By the end of Motion Graphs & Kinematic Equations you should be able to:
- I can read velocity from the slope of a position-time graph and acceleration from the slope of a velocity-time graph.
- I can find displacement from the area under a velocity-time graph.
- I can select and apply the kinematic equations to solve constant-acceleration problems.
6 questions, numeric and multiple choice, each with a worked explanation.
Vocabulary used: position-time graph, velocity-time graph, slope, area under the curve, kinematic equations.
Problem 1
Starting from rest, a cart accelerates at 5.0 m/s² for 3.0 s. Find its final velocity.
Answer and explanation
15 m/s
v = v₀ + at = 0 + (5.0)(3.0) = 15 m/s.
Problem 2
A velocity-time graph shows an object at a constant 8 m/s for 6.0 s. What displacement occurred?
Answer and explanation
48 m
Displacement is the area under the graph: 8 × 6.0 = 48 m.
Problem 3
Starting from rest, an object accelerates at 2.0 m/s² for 5.0 s. How far does it travel?
Answer and explanation
25 m
Δx = v₀t + ½at² = 0 + ½(2.0)(5.0²) = ½(2.0)(25) = 25 m.
Problem 4
On a position-time graph, a flat horizontal line means the object is…
- moving backward
- moving at constant velocity
- at rest
- accelerating
Answer and explanation
at rest
Zero slope means zero velocity, so the object's position isn't changing — it is at rest.
Problem 5
A cyclist moving at 15 m/s brakes at a constant −3.0 m/s². How far does she travel before stopping?
Answer and explanation
37.5 m
Use v² = v₀² + 2aΔx with v = 0: 0 = 15² + 2(−3.0)Δx, so 225 = 6Δx and Δx = 37.5 m.
Problem 6
On a velocity-time graph, the area between the line and the time axis represents the object's…
- jerk
- acceleration
- velocity
- displacement
Answer and explanation
displacement
Velocity multiplied by time gives displacement, so the area under a velocity-time graph is the displacement.