Explain the difference between distance and displacement.
Explain the difference between speed and velocity.
Define acceleration and state whether it is a scalar or a vector.
A cyclist rides 300 m east then 100 m west, taking 25 s. Find the distance, the displacement, the average speed and the average velocity.
A car speeds up from 5.0 m s−1 to 25 m s−1 in 8.0 s. Find its acceleration.
A train slows from 30 m s−1 to 12 m s−1 in 6.0 s. Find its acceleration and comment on the sign.
A displacement-time graph is a straight line from the origin to the point (10 s, 40 m). Find the velocity.
A velocity-time graph rises from 0 to 20 m s−1 in 4.0 s, then stays at 20 m s−1 for 6.0 s. Find the acceleration in the first phase and the total displacement.
On a curved displacement-time graph, a tangent at one point passes through (2.0 s, 10 m) and (6.0 s, 30 m). Find the instantaneous velocity at that point.
Explain why the distance travelled is never less than the magnitude of the displacement.
State what a zero gradient means on a displacement-time graph and on a velocity-time graph.
Total: 27 marks. Original work by the TheLucidSTEM team. Written in the style of the papers; no past paper question is reproduced.
Answer key · full worked solutionsclick to reveal
distance is the total path length travelled (a scalar); displacement is the straight-line distance and direction from start to finish (a vector).
speed is distance per unit time (a scalar); velocity is displacement per unit time and needs a direction (a vector).
acceleration is the rate of change of velocity, a = Δv / Δt; it is a vector.
distance = 400 m; displacement = 200 m east; average speed = 400 / 25 = 16 m s−1; average velocity = 200 / 25 = 8.0 m s−1 east.
a = (25 − 5.0) / 8.0 = 20 / 8.0 = 2.5 m s−2.
a = (12 − 30) / 6.0 = −18 / 6.0 = −3.0 m s−2. The negative sign shows the acceleration is opposite to the motion, so the train is slowing down.
velocity = gradient = 40 / 10 = 4.0 m s−1.
acceleration (first phase) = 20 / 4.0 = 5.0 m s−2. Displacement = ½ × 4.0 × 20 + 20 × 6.0 = 40 + 120 = 160 m.
instantaneous velocity = gradient of the tangent = (30 − 10) / (6.0 − 2.0) = 20 / 4.0 = 5.0 m s−1.
distance is the whole path length, while displacement is only the straight-line gap between start and finish. Any change of direction adds to the path but not to the straight-line gap, so the distance is greater than or equal to the magnitude of the displacement.
on a displacement-time graph a zero gradient means the velocity is zero (at rest); on a velocity-time graph a zero gradient means the acceleration is zero (constant velocity).