Difficulty: Intermediate | Prerequisites: Calculus II (integration), basic physics (Newton's laws, kinematics), vector operations (dot product, cross product, magnitude).
This topic extends the 1D and 2D kinematics you learned in introductory physics into full three-dimensional space using vector calculus. Instead of working with scalar equations of motion, you now describe acceleration, velocity, and position as vector-valued functions of time. The core technique is the same one you already know (integrate acceleration to get velocity, integrate velocity to get position) but each component of the vector is handled separately. If you are comfortable with basic integration and Newton's second law, this is a natural next step.
When forces act in three dimensions, break everything into component vectors. Integrate the constant acceleration vector once to get velocity (adding initial velocity as constants of integration), then integrate again to get position (adding initial position). To find when and where a projectile lands, set the height component to zero and solve the resulting quadratic.
Acceleration vector, a(t)
The vector whose components give the rate of change of velocity in each coordinate direction. In problems with only gravity and a constant wind force, a(t) is itself constant. Think of it as: the "push" acting on the object at every instant, split into x, y, and z directions.
Velocity vector, v(t)
The first integral of the acceleration vector. Each component tells you how fast the object is moving in that direction at time t. In simple terms, this means: integrate the acceleration once and add the initial velocity components as your constants of integration.
Position vector, s(t)
The second integral of the acceleration vector (first integral of velocity). Each component gives the object's coordinate at time t. Think of it as: the running total of where the object has travelled, starting from its initial position.
Initial conditions
The known values of velocity and position at t = 0. These determine the constants of integration. Without them, you cannot pin down a unique trajectory. In simple terms, this means: plug in t = 0 and match the result to whatever starting velocity and starting position the problem gives you.
Newton's second law (F = ma)
Force equals mass times acceleration. To find the acceleration caused by a force, divide the force vector by the object's mass. Think of it as: the bridge between a force (in newtons) and the acceleration (in m/s^2) it produces.
Speed at impact
The magnitude of the velocity vector at the moment of landing: |v(t)|. This is a scalar, not a vector. In simple terms, this means: compute each velocity component at the landing time, then use the Pythagorean theorem in 3D to get a single number.
Identify every force acting on the object: gravity, wind, thrust, etc.
Gravity always contributes -9.8 m/s^2 in the vertical (z) component.
For any other constant force, use F = ma to convert it to an acceleration: divide the force magnitude by the object's mass, then resolve into x, y, z components using the stated direction.
The total acceleration vector is the sum of all individual acceleration vectors.
Example: a 4 N wind blowing southeast (30 degrees east of true south) on a 0.5 kg rock gives a wind acceleration of 4/0.5 = 8 m/s^2. Decomposing: the x-component (east) is 8 sin(30) = 4, the y-component (south, taken as positive here depending on your coordinate system) is 8 cos(30) = 4 sqrt(3). Combined with gravity: a(t) = <4 sqrt(3), 4, -9.8>.
Integrate each component of a(t) with respect to t to get v(t), picking up one constant of integration per component.
Apply the initial velocity condition v(0) =
Example: if the rock is thrown northward at 45 degrees elevation with speed 34 m/s, the initial velocity in the northward direction is 34 cos(45) = 17 sqrt(2), the vertical component is 34 sin(45) = 17 sqrt(2), and the east-west component is 0. Depending on your axes: v(0) = <-17 sqrt(2), 0, 17 sqrt(2)>.
Result: v(t) = <(acceleration_x)t + v_x0, (acceleration_y)t + v_y0, (acceleration_z)t + v_z0>.
Integrate each component of v(t) with respect to t to get s(t), again picking up one constant of integration per component.
Apply the initial position condition s(0) =
Example: throwing from the top of a 100 m building at the origin gives s(0) = <0, 0, 100>.
Result: s(t) = <(1/2)(acceleration_x)t^2 + v_x0 t + x_0, ...> and so on for each component.
Set the z-component of s(t) equal to zero (ground level) and solve for t.
This is a quadratic equation in t. Use the quadratic formula.
You will get two solutions. Discard the negative one; the positive root is your landing time.
Example: -4.9t^2 + 17 sqrt(2) t + 100 = 0 gives t approximately equal to 7.594 seconds.
Plug the landing time back into s(t) to get the x and y coordinates of the landing point.
Plug the landing time into v(t) to get the velocity vector at impact.
The speed at impact is |v(t_land)| = sqrt(v_x^2 + v_y^2 + v_z^2).
Example: s(7.594) is approximately <17.2, 115.3, 0>, and the speed is approximately 65.4 m/s.
\vec{a}(t) = \langle a_x,\, a_y,\, a_z \rangle\vec{v}(t) = \langle a_x t + v_{x0},\, a_y t + v_{y0},\, a_z t + v_{z0} \rangle\vec{s}(t) = \left\langle \tfrac{1}{2}a_x t^2 + v_{x0}t + x_0,\; \tfrac{1}{2}a_y t^2 + v_{y0}t + y_0,\; \tfrac{1}{2}a_z t^2 + v_{z0}t + z_0 \right\rangleTo find landing time, solve the z-component for t = 0:
\tfrac{1}{2}a_z\, t^2 + v_{z0}\, t + z_0 = 0Speed at impact:
|\vec{v}(t_{\text{land}})| = \sqrt{v_x(t)^2 + v_y(t)^2 + v_z(t)^2}This is exactly how engineers model the trajectory of anything from a thrown ball to a launched satellite when forces beyond gravity are present. Wind-corrected ballistic calculations, sports analytics (e.g. modelling a football's path with crosswind), and robotics (predicting where a tossed object will land) all use this framework.
Students often forget to divide force by mass before adding it to the acceleration vector. The acceleration from a 4 N force on a 0.5 kg object is 8 m/s^2, not 4 m/s^2.
Students sometimes add initial conditions to the acceleration vector instead of to the velocity and position vectors. The constants of integration appear only when you integrate.
Students confuse speed (a scalar, the magnitude of the velocity vector) with the velocity vector itself. The exam may ask for speed, and giving a vector will lose marks.
Students occasionally forget that a quadratic has two roots and submit the negative time as an answer. Always discard the physically meaningless root.
This type of problem is a staple of Calculus III exams. Expect a full worked problem: given a scenario, find acceleration, velocity, position, landing time, landing position, and impact speed.
Pay close attention to coordinate system setup. The notes use a system where one horizontal direction is "northward" and another is "eastward," with z pointing up. Your professor may define axes differently.
The quadratic formula step is where arithmetic errors most commonly occur. Double-check your discriminant.
True or False: If the only force on an object is gravity, the acceleration vector is <0, 0, -9.8>. True.
Fill in the blank: To find velocity from acceleration, you ______ the acceleration vector and add ______. Integrate; the initial velocity (constants of integration).
True or False: The speed at impact is the velocity vector evaluated at the landing time. False. Speed is the magnitude of that vector.
Fill in the blank: To find when a projectile hits the ground, set the ______ component of the position vector equal to zero. z (vertical).
True or False: Both roots of the quadratic for landing time are physically meaningful. False. Discard the negative root.
Q: A 2 kg ball is thrown from the origin with initial velocity <10, 0, 20> m/s. The only force is gravity. Write the position vector s(t).
A: a(t) = <0, 0, -9.8>. Integrate: v(t) = <10, 0, -9.8t + 20>. Integrate again with s(0) = <0, 0, 0>: s(t) = <10t, 0, -4.9t^2 + 20t>.
Q: Using the position vector from the previous question, at what time does the ball hit the ground?
A: Set -4.9t^2 + 20t = 0. Factor: t(-4.9t + 20) = 0. So t = 0 (launch) or t = 20/4.9, which is approximately 4.08 seconds.
Q: A constant horizontal wind exerts a force of 6 N on a 3 kg object. What acceleration does this add?
A: a = F/m = 6/3 = 2 m/s^2 in the direction of the wind. If the wind blows due east, this adds 2 m/s^2 to the x-component of the acceleration vector.
Q: Explain why the acceleration vector in the class example is constant even though the object is moving.
A: Both gravity and the wind force are constant (they do not depend on position or time). A constant force on a constant mass produces a constant acceleration by F = ma. The object's velocity and position change, but the acceleration does not.
Q: After finding the landing time t = 7.594 s for the class example, how do you compute the impact speed?
A: Evaluate each component of v(t) at t = 7.594, then compute the magnitude: |v(7.594)| = sqrt(v_x^2 + v_y^2 + v_z^2), which is approximately 65.4 m/s.
This connects directly to the TNB (tangent, normal, binormal) frame material earlier in MATH 163, where you decompose acceleration into tangential and normal components using curvature. Both topics rely on differentiating and integrating vector-valued functions of time.
Domain-finding for multivariable functions (Section 14.1) is the next major topic in the course. While that section deals with functions of the form f(x, y) rather than vector-valued functions of t, the underlying skill of working comfortably in three-dimensional coordinate space carries over.
The integration techniques here also preview the ideas behind line integrals and work integrals later in the course.
3D kinematics, vector-valued functions, projectile motion with wind, Newton's second law in 3D, constant acceleration vector, integrating acceleration, velocity from acceleration, position from velocity, landing time quadratic, impact speed, magnitude of velocity, calculus III physics, MATH 163, multivariable calculus kinematics, 3D trajectory, vector integration, initial conditions, constants of integration