Learning objectives
Distinguish distance from displacement and speed from velocity, and determine average and instantaneous values from tables, graphs and equations of motion
Interpret displacement–time, velocity–time and acceleration–time graphs using gradients and areas under the curve
Apply the four equations of uniformly accelerated motion to straight-line problems, including free fall with g = 9.81 m s⁻²
Resolve projectile motion into independent horizontal and vertical components, and describe qualitatively how fluid resistance changes the motion
Key definitions
Displacement
The straight-line vector from the starting point to the finishing point of a motion, measured in metres. Its magnitude can never exceed the distance travelled.
Instantaneous velocity
The rate of change of displacement at one instant, equal to the gradient of the tangent to a displacement–time graph at that point.
Uniform acceleration
Motion in which the rate of change of velocity is constant, so that a velocity–time graph is a straight line.
Free fall
Motion under the action of the gravitational force alone, with fluid resistance neglected, giving a downward acceleration of magnitude g = 9.81 m s⁻² near the Earth.
Terminal speed
The constant speed reached by a falling body when the upward drag force has grown until it balances the weight, so the resultant force and the acceleration are zero.
Projectile
A body given an initial velocity and then moving freely under gravity, with a constant horizontal velocity component and a uniformly changing vertical component.
Explanation
Scalars and vectors. Distance is the total path length travelled and is a scalar; displacement is the vector from start to finish. An athlete who runs one complete lap of a 400 m track covers a distance of 400 m but has zero displacement, so the average speed is 400 m divided by the time while the average velocity is zero. Average speed is total distance divided by total time; average velocity is total displacement divided by total time. Instantaneous values refer to a single instant and are found from the gradient of a tangent. In IB answers keep the vector language consistent: a velocity of −6.0 m s⁻¹ is not "a negative speed", it is a speed of 6.0 m s⁻¹ in the direction chosen as negative.
Reading motion graphs. On a displacement–time graph the gradient is the velocity: a horizontal line means the body is at rest, a straight sloping line means constant velocity, and a curve means the velocity is changing. On a velocity–time graph the gradient is the acceleration and the area between the line and the time axis is the displacement, with area below the axis counted as negative. On an acceleration–time graph the area gives the change in velocity. A common Paper 1B task is to take a gradient using a large triangle drawn on the printed graph, quoting the coordinates of both corners; a small triangle magnifies the reading uncertainty and loses marks.
The equations of uniformly accelerated motion. When the acceleration is constant, v = u + at, s = ut + ½at², v² = u² + 2as and s = ½(u + v)t. Each equation leaves out one quantity, so choose the one that omits the variable you neither know nor want. Set up a sign convention first and keep it: if upwards is positive then a = −9.81 m s⁻² for a body in free fall, and a stone thrown upwards has positive u and negative displacement once it falls below the launch point. At the highest point of a vertical flight the velocity is zero but the acceleration is still 9.81 m s⁻² downwards — that is what makes the body come back.
Projectiles. With fluid resistance neglected, the horizontal and vertical motions of a projectile are independent: the only force is the weight, which acts vertically, so the horizontal component of velocity is constant while the vertical component changes at 9.81 m s⁻² downwards. Resolve the launch velocity first: ux = u cosθ and uy = u sinθ. Time is the quantity shared by the two motions, so nearly every projectile problem is solved by finding the time from the vertical motion and then substituting it into x = uxt. At the top of the path vy = 0, so the time to the top is u sinθ/g; for a launch and landing at the same height the total time of flight is twice that and the range is R = u² sin(2θ)/g, which is greatest at θ = 45°. The velocity at any instant is the vector sum of ux and vy, so its magnitude is √(ux² + vy²) and its direction is found from tanθ = vy/ux.
Fluid resistance. Real projectiles move through air, which exerts a drag force opposite to the velocity whose magnitude increases with speed. The path is no longer a symmetrical parabola: the range, the maximum height and the time of flight are all reduced, the descent is steeper than the ascent, and the speed on landing is smaller than the launch speed. For a body falling from rest the weight is at first much larger than the drag, so the acceleration starts at g and then falls as the drag grows. When the drag force has grown to equal the weight the resultant force is zero, the acceleration is zero and the body continues at a constant terminal speed. Note carefully that the body is still moving and still has the weight acting on it; it is the resultant force, not the force, that has become zero.
Finding an acceleration experimentally. A standard investigation releases a trolley or a ball from rest and records the distance s fallen or travelled in a time t, using light gates, a motion sensor or video frames. Since s = ½at² for a start from rest, a graph of s against t² should be a straight line through the origin of gradient ½a, so a is twice the gradient. Plotting t² rather than t is the step that linearises the relationship and makes the gradient meaningful. To estimate the uncertainty, draw the steepest and the shallowest lines that still pass through all the error bars and take half the difference of their gradients as the absolute uncertainty in the gradient. A systematic delay in starting the timer shifts the whole line and shows up as a non-zero intercept, while random timing scatter shows up as points spread about the line.
Formulae
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t
Projectile: x = u cosθ × t and y = u sinθ × t − ½gt²
Range on level ground: R = u² sin(2θ)/g
g = 9.81 m s⁻² near the Earth surface
Worked examples
q: A ball is launched from level ground at 25 m s⁻¹ at 35° above the horizontal. Air resistance is negligible. Determine the time of flight, the horizontal range and the maximum height reached.
steps: Resolve the launch velocity: horizontal component = 25 cos 35° = 20.48 m s⁻¹, vertical component = 25 sin 35° = 14.34 m s⁻¹.
Take upwards as positive, so a = −9.81 m s⁻². At the top the vertical velocity is zero, so the time to the top is t = 14.34/9.81 = 1.462 s.
Launch and landing are at the same height, so the time of flight is 2 × 1.462 = 2.92 s.
Horizontal velocity is constant, so the range is 20.48 × 2.92 = 59.9 m (3 significant figures).
Maximum height from v² = u² + 2as with v = 0: h = 14.34²/(2 × 9.81) = 10.5 m.
In context
Vehicle safety testing uses exactly this analysis: stopping distance grows with the square of the speed, because v² = u² + 2as makes the braking distance proportional to u² for a fixed deceleration.
Skydivers exploit terminal speed. A diver in a head-down position reaches about 90 m s⁻¹, but opening a parachute multiplies the drag area so the drag briefly exceeds the weight, the diver decelerates, and a new and far smaller terminal speed is reached.
Common misconceptions
Students often say that the acceleration of a ball is zero at the top of its flight because the velocity is zero. In fact the only force is still the weight, so the acceleration is still 9.81 m s⁻² downwards.
Students often treat the horizontal and vertical motions of a projectile as linked. In fact with no air resistance the horizontal velocity is constant and the vertical motion is the same as free fall, and the two are connected only through the shared time.
Students often state that a body at terminal speed has no force acting on it. In fact the weight and the drag are both large; it is their resultant that is zero.
Students often read the area under a displacement–time graph as though it meant something. In fact only the gradient of that graph is physically meaningful; areas are read from velocity–time and acceleration–time graphs.
Study tips
Write down a sign convention before any free-fall calculation and state it in your answer, then keep every substituted value consistent with it.
In projectile questions set out two columns, one horizontal and one vertical, and put the time in both. This makes the shared variable obvious to you and to the examiner.
When a question says "determine" a gradient from a printed graph, use a triangle spanning at least half the plotted line and write down the coordinates you used, then show the subtraction.
Carry unrounded values through the working and round only the final answer, quoting it to the same number of significant figures as the least precise datum, usually two or three.
Reasoning
The equations of motion here describe how a body moves; the next topic, forces and Newton laws, explains why it moves that way.
Summary
Distance and speed are scalars; displacement, velocity and acceleration are vectors and need a stated sign convention.
On a displacement–time graph the gradient is velocity; on a velocity–time graph the gradient is acceleration and the area is displacement.
The four equations of motion apply only when the acceleration is constant.
Projectile motion separates into constant horizontal velocity and free-fall vertical motion, linked only by the time.
Terminal speed is reached when the drag force has grown to balance the weight, so the resultant force is zero.
Check your understanding
Question 1: A car travels 60 m north then 80 m south in 20 s. Its average velocity is
1. 1.0 m s⁻¹ south
2. 7.0 m s⁻¹ south
3. 1.0 m s⁻¹ north
Question 2: The area between a velocity–time graph and the time axis gives
1. the acceleration
2. the average speed
3. the displacement
Question 3: For a projectile with air resistance neglected, the horizontal component of velocity
1. is largest at the top of the path
2. stays constant throughout the flight
3. decreases steadily to zero