Scalars and vectors
Vector — A quantity with both magnitude and direction. A scalar has magnitude only.
Some quantities are fully described by a number and a unit: mass, time, temperature, energy, speed. These are scalars, and they add like ordinary numbers — 3 kg of rice plus 2 kg of rice is 5 kg however you arrange it.
Others need a direction as well: displacement, velocity, acceleration, force, momentum. These are vectors, and they do not add like ordinary numbers. Walk 3 km east and then 4 km north and you have travelled 7 km, but you are only 5 km from where you started.
That difference is the whole of this topic. Distance is the path length, a scalar. Displacement is the straight line from start to finish, a vector. Speed is a scalar; velocity is speed with a direction attached. A car going round a roundabout at a steady 30 km/h has constant speed and continuously changing velocity.
| Scalar | Vector |
|---|---|
| distance | displacement |
| speed | velocity |
| mass | weight |
| energy, work, power | force, momentum |
| time, temperature | acceleration |
A direction needs a reference
Never write a direction as a bare angle. "36.9°" means nothing on its own; "36.9° north of east" or "36.9° above the horizontal" is an answer. Mark schemes routinely award the magnitude and withhold the direction mark for exactly this.
Adding vectors
Two vectors are added by placing them nose to tail: draw the first, start the second where the first ended, and the resultant runs from the very beginning to the very end. Drawn to scale on graph paper this gives an answer good enough for many questions.
The parallelogram method is the same idea drawn differently — both vectors from a common origin, complete the parallelogram, and the resultant is the diagonal.
When the two vectors are at right angles, no drawing is needed. Pythagoras gives the magnitude and the tangent gives the direction, and this is by far the most common case in an exam.
When they are at some other angle, either resolve both into components and add the components, or use the cosine rule. Resolving is more reliable, because it turns any number of vectors into two simple sums.
- R
- resultant magnitude
- Rₓ
- sum of the x components
- R_y
- sum of the y components
- θ
- angle from the x axis°
Change either arrow and watch the resultant. Set both to the same angle and the magnitudes simply add; set them 180° apart and they subtract; set them at 90° and the resultant is the hypotenuse. Every case in the topic is somewhere on those sliders.
Resolving into components
Any vector can be replaced by two perpendicular vectors that together have exactly the same effect. Those are its components, and replacing a vector by them is called resolving.
The component along the direction you are measuring from uses cosine; the component at right angles to it uses sine. Which is which depends on where the angle is measured from, so it is worth checking rather than memorising: if the angle is measured from the x axis, then Aₓ = A cos θ.
A sanity check that never fails: as the angle goes to zero the vector should lie entirely along x, so the x component should equal the full magnitude. Substitute 0° and see whether your formula gives that.
Once every vector in a problem is in components, adding them is just two ordinary sums — all the x components, then all the y components. Recombine at the end with Pythagoras.
- A
- magnitude
- θ
- angle from the x axis°
Three forces act on a point: 12 N east, 8.0 N north, and 6.0 N at 30° north of west. Find the resultant.
- Resolve the third force:
x = −6.0 cos 30° = −5.20 N,y = +6.0 sin 30° = +3.00 N.West is negative x, so the sign is part of the answer. - Sum the x components:
12 + 0 − 5.20 = 6.80 N. - Sum the y components:
0 + 8.0 + 3.00 = 11.0 N. R = √(6.80² + 11.0²) = √(46.2 + 121) = 12.9 N.θ = tan⁻¹(11.0 / 6.80) = 58.3°.- So 12.9 N at 58.3° north of east.The direction needs a reference to mean anything.
12.9 N at 58.3° north of east
Equilibrium
Equilibrium — The state of a body with no resultant force and no resultant moment acting on it, so it stays at rest or continues at constant velocity.
A body in equilibrium has forces that cancel completely. In component form that means the x components sum to zero and the y components sum to zero — two separate equations, and both must be written.
For a body acted on by exactly three forces there is a neat geometric consequence: if the three are drawn nose to tail they form a closed triangle, because the resultant is zero. That triangle can be drawn to scale and measured, which is often quicker than the algebra.
Rotation must also be considered. A ladder against a wall can have zero resultant force and still slide, so the second condition — that the moments about any point balance — is needed too.
Key points
- Scalars have magnitude; vectors have magnitude and direction.
- Perpendicular vectors combine by Pythagoras; the angle comes from the tangent.
- Resolving turns any set of vectors into two ordinary sums.
- A direction is only an answer when it names a reference — "north of east", not "58°".
- Equilibrium needs
ΣFₓ = 0,ΣF_y = 0andΣτ = 0.
When the acceleration will not stay still
The suvat equations carry one hidden condition that is easy to forget: the acceleration must be constant. Apply them to a car whose acceleration is changing and every answer will be wrong, with no warning.
When acceleration varies, calculus replaces the formulas. Differentiating moves down the chain from position to velocity to acceleration, because each is the rate of change of the one before. Integrating moves back up, and each integration introduces a constant that an initial condition must pin down.
- s
- displacementposition measured from a stated origin
- v
- velocitythe gradient of the displacement-time graph
- a
- accelerationthe gradient of the velocity-time graph
A particle moves in a straight line so that its velocity is v = 3t² − 12t + 9 m s⁻¹. It starts at the origin. Find (a) when it is instantaneously at rest, (b) its acceleration at t = 1, (c) its displacement at t = 4.
- (a) At rest means v = 0: 3t² − 12t + 9 = 0, so t² − 4t + 3 = 0.Instantaneously at rest means the velocity is zero, not the acceleration.
- (t − 1)(t − 3) = 0, so t = 1 s and t = 3 s.Two separate moments — the particle stops, reverses, and stops again.
- (b) a = dv/dt = 6t − 12, so at t = 1, a = −6 m s⁻².Differentiating velocity gives acceleration. The negative sign means it is decelerating in the positive direction.
- (c) s = ∫(3t² − 12t + 9)dt = t³ − 6t² + 9t + c.Integrating velocity gives displacement, with a constant to be determined.
- Starts at the origin, so s = 0 when t = 0, giving c = 0.The initial condition fixes the constant. Without it the answer is a family of solutions, not a number.
- s = 64 − 96 + 36 = 4 m.Displacement, not distance — the particle reversed direction twice, so the distance travelled is larger.
(a) t = 1 s and t = 3 s; (b) −6 m s⁻²; (c) 4 m
Displacement is not distance
If the velocity changes sign during the interval, the particle doubles back and the two contributions partly cancel. ∫v dt gives the displacement. For the distance travelled you must split the integral at each time where v = 0 and add the magnitudes of the separate pieces. A question asking for distance when the velocity changes sign is testing exactly this.