Understanding Motion
This guide explains the why behind every kinematics concept — from what displacement really means to how to attack projectile problems systematically. Work through each section, try every checkpoint, then reveal the answer.
When you travel somewhere, two different measurements describe your journey: how far you physically moved (distance), and where you ended up relative to where you started (displacement). These are often different — sometimes very different.
A runner who completes 4 laps of a 400 m track has run a distance of 1600 m. But their displacement is exactly 0 m — they are standing at the same point they started. If you want to know their average velocity, you use displacement (0 m), not distance (1600 m). Average velocity would be 0 m/s. Average speed, however, would be 1600 m divided by the total time — a very different number.
Displacement: Δd = d_f − d_i
Average velocity: v_avg = Δd / Δt
Average speed: speed = total distance / total time
Always define your positive direction first. Every sign in the problem depends on it.
Δd = +300 m, Δt = 12 s
Since the result is positive and north is our + direction, the velocity is northward.
speed = 700 / 35 = 20 m/s
b) A runner moves at v = 18 m/s for 4.5 s. How far does she travel?
c) A person walks 5 km east, then 5 km west in 2 hours total. What is their average velocity? What is their average speed?
a) v = Δd / Δt = 150 / 6 = 25 m/s
b) Δd = v × Δt = 18 × 4.5 = 81 m
c) Displacement = 5 − 5 = 0 km. Average velocity = 0 / 2 h = 0 km/h. Total distance = 10 km. Average speed = 10 / 2 = 5 km/h. Same time, very different answers — because velocity uses displacement and speed uses distance.
Acceleration is one of the most misunderstood concepts in kinematics. Most students think "acceleration means speeding up." But acceleration means any change in velocity — including slowing down, changing direction, or both at once.
Acceleration measures the rate of change of velocity. If your velocity is changing — whether it is increasing, decreasing, or shifting direction — you are accelerating. A car braking from 30 m/s to 0 m/s has its velocity changing, so it has acceleration (a negative one, since it opposes the direction of motion). The word "deceleration" just means acceleration in the direction opposite to the object's motion. Physics uses one unified concept.
If the result is positive: acceleration is in the positive direction.
If the result is negative: acceleration is in the negative direction (opposing the + direction).
Units: m/s² (metres per second squared — velocity changing by X m/s every second).
v_f = 27 m/s
t = 9 s
v_f = +6 m/s
t = 4 s
The negative sign means the acceleration points backward (opposing forward motion). This is deceleration — the car is slowing down.
b) A car has a = −4 m/s² and starts at v₀ = 24 m/s. How long until it stops?
c) Explain why a car can be "accelerating" while slowing down.
a) a = (20 − 5) / 5 = 15 / 5 = 3 m/s²
b) v_f = 0 = v₀ + at → 0 = 24 + (−4)t → 4t = 24 → t = 6 s
c) Acceleration means the velocity is changing. When a car slows down, its velocity decreases — that is a change in velocity. Since a = Δv/Δt and Δv is non-zero, the car has acceleration (negative in this case, opposing its direction of travel). "Decelerating" is simply a word for negative acceleration — it is still acceleration in the physics sense.
When acceleration is constant, five kinematic equations connect the five variables: displacement (d), initial velocity (v₀), final velocity (v), acceleration (a), and time (t). In any problem you will know three of these and need to find one or two others.
Each equation links a different combination of four of the five variables. The trick is to identify which variable you are missing, then pick the equation that does not include it. For example, if you are not given time (t) and don't need it, use Equation 3 (v² = v₀² + 2ad) — time never appears in it.
1. v = v₀ + at
2. d = v₀t + ½at²
3. v² = v₀² + 2ad
4. d = ((v₀ + v) / 2) × t
5. d = vt − ½at²
These ONLY work when acceleration is constant.
Find: d
d = 0 + ½(2.5)(64)
d = ½ × 160 = 80 m
Find: a (t is unknown and not needed)
0 = 625 + 125a
125a = −625
a = −5 m/s²
Negative acceleration means the car decelerates — makes sense, the car is slowing down.
b) An object is dropped from 45 m. How long does it take to land? (v₀ = 0, a = 9.8 m/s²)
c) Which UAM equation would you use if time (t) is unknown and not needed?
a) v = v₀ + at = 10 + 3(5) = 25 m/s
d = v₀t + ½at² = 10(5) + ½(3)(25) = 50 + 37.5 = 87.5 m
b) d = v₀t + ½at² → 45 = 0 + ½(9.8)t² → 45 = 4.9t² → t² = 9.18 → t ≈ 3.03 s
c) Equation 3: v² = v₀² + 2ad — this equation contains no t, so it is used whenever time is the missing/unwanted variable.
Motion graphs are powerful tools — each graph encodes kinematic information in its shape, slope, and the area beneath it. Mastering graphs means you can extract velocity, acceleration, and displacement from a visual without any equations.
Slope = rise/run = (change in y) / (change in x). On a d-t graph, that is Δd/Δt — the definition of velocity. On a v-t graph, it is Δv/Δt — the definition of acceleration. The graphs are connected: slope of d-t gives v-t; slope of v-t gives a-t. And area runs the other way: area under a-t gives change in velocity; area under v-t gives displacement.
d-t graph: slope = velocity
v-t graph: slope = acceleration, area under curve = displacement
a-t graph: area under curve = change in velocity (Δv)
| Graph | Shape | Means |
|---|---|---|
| d-t | Straight line, positive slope | Constant positive velocity |
| d-t | Horizontal line | Object is stationary (v = 0) |
| d-t | Curve (steepening) | Accelerating (velocity increasing) |
| v-t | Straight line, positive slope | Constant positive acceleration |
| v-t | Horizontal line | Constant velocity (a = 0) |
| v-t | Line crossing x-axis | Object reverses direction |
| a-t | Horizontal line | Uniform (constant) acceleration |
Constant 15 m/s for 6 s forms a rectangle on the v-t graph.
b) A v-t graph is flat at 12 m/s for 3 s, then drops linearly to 0 m/s over 2 s. What is the total displacement?
c) What does a negative slope on a v-t graph tell you about the motion?
a) On a d-t graph, slope = velocity = 8 m/s. Displacement = 8 × 4 = 32 m (or read from the graph directly as change in d).
b) Rectangle: 12 × 3 = 36 m. Triangle: ½ × 2 × 12 = 12 m. Total = 36 + 12 = 48 m
c) A negative slope on a v-t graph means the acceleration is negative — the object is decelerating (slowing down if moving in the positive direction, or speeding up in the negative direction). If the line crosses the x-axis, the object momentarily stops and then moves in the opposite direction.
A projectile is any object that has been given an initial velocity and then moves freely under gravity. The genius of solving these problems lies in one key insight: the horizontal and vertical motions are completely independent of each other. Gravity only pulls down — it has no effect on horizontal motion whatsoever.
Forces only affect motion in their own direction. Gravity pulls straight down, so it only accelerates the object downward — it cannot accelerate it sideways. With no horizontal force, horizontal velocity stays exactly constant throughout the flight. This is why a bullet fired horizontally and a bullet simply dropped from the same height hit the ground at the same time — they share identical vertical motion.
Horizontal: aₓ = 0, vₓ = constant = v₀ cos θ, x = vₓ · t
Vertical: aᵧ = −9.8 m/s² (up is +), vᵧ₀ = v₀ sin θ, vᵧ = vᵧ₀ − gt
The time of flight calculated from vertical equations is the same t used in horizontal equations.
Vertical drop: d = 20 m (downward, so +20 m if down is +)
20 = 0 + ½(9.8)t²
20 = 4.9t²
t² = 20/4.9 = 4.082
t ≈ 2.02 s
vᵧ₀ = 25 sin 37° ≈ 25 × 0.602 ≈ 15 m/s
t_up = 15 / 9.8 ≈ 1.53 s
Total time = 2 × t_up ≈ 3.06 s
0 = 15² − 2(9.8)h
0 = 225 − 19.6h
h = 225 / 19.6 ≈ 11.5 m
b) A ball is launched at 30° above horizontal at 20 m/s. Find the maximum height reached.
c) Why does a horizontally launched ball and a ball simply dropped from the same height hit the ground at exactly the same time?
a) Vertical: 45 = ½(9.8)t² → t² = 45/4.9 = 9.18 → t ≈ 3.03 s
Range: x = 18 × 3.03 ≈ 54.5 m
b) vᵧ₀ = 20 sin 30° = 20 × 0.5 = 10 m/s
At peak: vᵧ = 0 → use v² = vᵧ₀² − 2gh → 0 = 100 − 19.6h → h = 100/19.6 ≈ 5.1 m
c) Horizontal and vertical motions are independent. Both balls have identical vertical motion: both start with vᵧ₀ = 0 (neither has an initial vertical velocity) and both experience the same gravitational acceleration downward. Since the vertical equations are identical for both, they take exactly the same time to fall the same vertical distance. The horizontal velocity of the launched ball has no effect on how quickly it falls.