AP PHYSICS C: MECHANICS › UNIT 1, KINEMATICS › TOPIC 1.5
Motion in Two or Three Dimensions
Every derivative in this unit has been taken with respect to time, and none of them cared how many directions there were. This topic collects the position, the velocity and the acceleration into single vector objects, and the calculus comes through untouched.
1.5.A Write the position of an object as a vector function of time, and obtain the velocity and the acceleration by differentiating it component by component.
1.5.B Describe motion in two or three dimensions by treating each component as an independent one-dimensional problem joined to the others only by the shared clock.
1. One object, one vector function, one variable
Bold letters are vectors, and i, j, k are unit vectors along the three axes. The position of a particle is then a single function of a single variable:
r(t) = x(t)i + y(t)j + z(t)k
This is not a new kind of function. It is three ordinary scalar functions of t carried around in one bracket. The unit vectors are constants, and that single fact is what makes everything below work.
2. Differentiate component by component
With the unit vectors constant, the derivative of the sum is the sum of the derivatives of the coefficients:
v(t) = dr/dt = (dx/dt)i + (dy/dt)j + (dz/dt)k
a(t) = dv/dt = (d²x/dt²)i + (d²y/dt²)j + (d²z/dt²)k
Everything from Topics 1.2 and 1.3 now applies once per axis. Integration runs the same way, one constant of integration per component, each fixed by its own initial condition:
v(t) = v0 + ∫a dtr(t) = r0 + ∫v dt
The components do not interact. A force applied along x changes vx and leaves vy alone. The only thing the components share is t, and every two-dimensional problem you will ever solve is solved by finding t from one component and spending it on the other.
3. What the two derivatives look like on the path
Drawn on the curve, the velocity and the acceleration behave very differently. v is always tangent to the path, because it is the limit of a chord as the chord shrinks. a is not tangent in general: it leans toward the inside of the bend, and only its tangential part changes the speed.
The particle runs along x = A cos t, y = B sin t. Blue is v, gold is a. Set the shape to a circle and the two arrows stay exactly perpendicular while the speed never changes. Flatten it into an ellipse and the angle between them stops being a right angle, and the speed starts to vary, at precisely the moments when a has a component along the direction of travel.
For this path a = −r, so the acceleration always points at the origin. On the circle that direction is perpendicular to the motion at every instant, which is the whole content of uniform circular motion: an acceleration that turns the velocity without lengthening it.
4. Projectile motion is one vector equation
Take the constant acceleration a = −gj and integrate it twice. Nothing about the vector form is harder than the scalar case:
v(t) = v0 − gtjr(t) = r0 + v0t − ½gt²j
Read off the components and the two familiar statements fall out: x = x0 + v0xt because there is no i term in the acceleration, and y = y0 + v0yt − ½gt², which is free fall. Eliminate t between them and the path is a parabola, which is a result rather than an assumption.
The blue arrow is the velocity, drawn tangent to the path, with its two components shown against it. The gold arrow is the acceleration, which never changes in size or direction for the whole flight. Watch the horizontal component of the velocity: it is the same number at the launch, at the apex, and at the landing.
5. Three dimensions cost nothing, and the frames still work
Adding a third component adds a third scalar function and a third initial condition. It adds no new physics, because the argument in section 2 never counted the axes. Everything from Topic 1.4 carries over the same way: the relative-velocity statement becomes vPA = vPB + vBA, a vector equation now, which means it is one equation per axis and each of those is the one-dimensional statement you already proved.
Two quantities are worth naming, because they are the ones a vector formulation makes easy to confuse. The speed is |v|, a positive scalar. The velocity is the vector, and it can change while the speed does not, which is what happens on every bend. A car going round a corner at a steady reading on the speedometer is accelerating the entire way round.
Check yourself
1. A particle has r(t) = (3t²)i + (4t − t²)j in meters. Find v(t) and a(t), and give the speed at t = 1 s.
v = (6t)i + (4 − 2t)j and a = 6i − 2j, a constant. At t = 1 the velocity is 6i + 2j, so the speed is √40 = 6.3 m/s. Note that the speed is not 6 + 2: perpendicular components never simply add.
2. For the same particle, at what time is the velocity purely horizontal, and what is the acceleration at that instant?
Set vy = 4 − 2t = 0, giving t = 2 s. The acceleration is still 6i − 2j, because it does not depend on t at all. A zero component of velocity says nothing whatever about the acceleration.
3. A projectile is launched at 20 m/s at 30° above the horizontal from ground level, with g = 10 m/s². Find the hang time and the range.
v0y = 20 sin 30° = 10 m/s, so y = 10t − 5t² = 0 gives t = 2.0 s. Then v0x = 20 cos 30° = 17.3 m/s and the range is 17.3 × 2.0 = 34.6 m. The time came from one component and was spent on the other, which is the whole method.
4. Explain, using the derivative and not a picture, why the velocity of a particle is always tangent to its path.
v is the limit of Δr/Δt. The numerator is the chord joining two points on the path, and the denominator is a positive scalar, so the ratio points along the chord. As Δt goes to zero the second point slides into the first and the chord becomes the tangent. Nothing in that argument mentions how many dimensions there are.
5. A particle moves so that |v| is constant. Show that a must be perpendicular to v.
Constant speed means v · v is constant. Differentiate: d(v · v)/dt = 2v · a = 0, so either the acceleration is zero or it is perpendicular to the velocity. This is the general version of what SIM 1 shows on the circle, and it is where circular motion in Unit 3 begins.
Unit 2, Force and Translational Dynamics. Unit 1 has described motion completely without ever asking what causes it. Unit 2 answers that, and it does so with a vector equation, resolved component by component in exactly the way this topic has just set up.