What is the difference between distance and displacement?

Distance is the total path length covered by an object, and it is always a positive scalar quantity. Displacement is the shortest straight-line change in position from the starting point to the final point, and it is a vector quantity (it has both magnitude and direction). Displacement can be zero even when distance is not — for example, when an object returns to its starting point — but the magnitude of displacement can never exceed the distance travelled.

What are the three equations of motion for constant acceleration, and what does each variable mean?

The three equations are v = u + at, s = ut + ½at², and v² = u² + 2as, where u is initial velocity, v is final velocity, a is acceleration, s is displacement, and t is time. They apply only when acceleration is constant, and each equation is useful depending on which three of the five quantities (u, v, a, s, t) are already known in a problem.

Why is an object in uniform circular motion said to be accelerating even though its speed doesn't change?

Acceleration is the rate of change of velocity, and velocity depends on both speed and direction. In uniform circular motion, the speed stays constant, but the direction of motion is continuously changing at every point along the circle, so the velocity is continuously changing — which means the object is always accelerating, even though its speedometer reading never changes.

How do you read velocity and displacement from motion graphs?

On a position-time graph, the slope of the line at any point gives the object's velocity at that instant — a steeper slope means higher velocity, and a horizontal (flat) line means the object is at rest. On a velocity-time graph, the slope gives acceleration, while the area enclosed between the graph line and the time axis gives the displacement over that time interval.

Why does doubling a vehicle's speed more than double its stopping distance?

Because braking distance depends on the square of the initial velocity (from v² = u² + 2as, with v = 0 at the stop: s = u²/2a). If the acceleration during braking stays the same, doubling the initial speed u means the stopping distance s becomes four times as large, not just twice — which is why high speeds are disproportionately more dangerous when a vehicle needs to stop suddenly.

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