Flashcards · 9702 · AS Topic 2

Kinematics (AS Level) flashcards

8 cards on AS Topic 2. Pick a subtopic or study them all.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 8 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

The four equations of motion, and the condition for using them: v = u + at s = ut + ½at² v² = u² + 2as s = (u + v)t ÷ 2 Valid only for uniform (constant) acceleration.
Derive v = u + at: Acceleration is the rate of change of velocity: a = (v − u) ÷ t. Rearranging: at = v − u, so v = u + at.
Velocity–time graph: what do the gradient and the area give?: Gradient: acceleration. Area under the line: displacement (area below the time axis counts as negative displacement).
How do you find the velocity at an instant from a curved displacement–time graph?: Draw the tangent to the curve at that instant and find its gradient (not the gradient of a chord between two points).
The key method for projectile motion: Treat horizontal and vertical motion independently; they share only the time. Horizontal: a = 0, constant velocity. Vertical: a = g downwards, use the equations of motion.
A ball is thrown at 20 m s⁻¹ at 30° above the horizontal. Find its initial velocity components.: Horizontal: 20 cos 30° = 17.3 m s⁻¹. Vertical: 20 sin 30° = 10 m s⁻¹.
At the top of a projectile’s path, which velocity component is zero?: The vertical component. The horizontal component is unchanged (with no air resistance).
How can g be determined by a free-fall experiment?: Drop an object from rest through a measured height s, timing the fall electronically (light gates or an electromagnet release). s = ½gt², so g = 2s ÷ t². Repeat and average t.