Physics

Kinematics

Study of motion without considering its causes. Position vector, displacement, velocity, and acceleration. Intrinsic components of acceleration. Equations for Uniform Linear Motion (ULM) and Uniformly Accelerated Linear Motion (UALM).

Kinematics is the branch of physics that describes the motion of points, bodies, and systems of bodies without considering the forces that cause them to move.
Position
Position Vector, r⃗(t)\vec{r}(t): indicates the position coordinates of the particle at each instant.
r⃗(t)=x(t)i^+y(t)j^\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}
r⃗(t)=(2t)i^+(4−t2)j^\htmlClass{text-primary}{\vec{r}(t) = (2t)\hat{i} + (4-t^2)\hat{j}}
Displacement Vector, Δr⃗\Delta\vec{r}: is the difference in position between two instants t1t_1 and t2t_2.
Δr⃗1→2=r⃗t2−r⃗t1\Delta\vec{r}_{1\to 2} = \vec{r}_{t_2} - \vec{r}_{t_1}
Δr⃗2→4=(8i^−12j^)−(4i^)=4i^−12j^\htmlClass{text-primary}{\Delta\vec{r}_{2\to 4} = (8\hat{i}-12\hat{j}) - (4\hat{i}) = 4\hat{i}-12\hat{j}}
Displacement ∣Δr⃗∣|\Delta\vec{r}|: linear distance between two points.
∣Δr⃗∣=∣r⃗t2−r⃗t1∣|\Delta\vec{r}| = |\vec{r}_{t_2} - \vec{r}_{t_1}|
∣Δr⃗2→4∣=42+122=12.65 m\htmlClass{text-primary}{|\Delta\vec{r}_{2\to 4}| = \sqrt{4^2 + 12^2} = 12.65\text{ m}}
Trajectory (Path): is the path followed by the moving object. Mathematically, it can be described as a parametric, implicit, or explicit equation.
{x=x(t)y=y(t)⇒y=f(x)\begin{cases} x=x(t) \\ y=y(t) \end{cases} \Rightarrow y=f(x)
{x=2ty=4−t2⇒y=4−x24\htmlClass{text-primary}{\begin{cases} x=2t \\ y=4-t^2 \end{cases} \Rightarrow y=4-\frac{x^2}{4}}
Distance traveled: Corresponds to the length of the path over which the body moves between two instants. It should not be confused with displacement.
Trajectory and distance traveled vs displacement
Velocity
Average velocity, v⃗avg\vec{v}_{avg}: is the ratio of displacement to the time interval.
v⃗avg=Δr⃗Δt=r⃗t2−r⃗t1t2−t1\vec{v}_{avg} = \frac{\Delta\vec{r}}{\Delta t} = \frac{\vec{r}_{t_2} - \vec{r}_{t_1}}{t_2 - t_1}
v⃗avg(2→4)=Δr⃗2→4Δt=4i^−12j^4−2=2i^−6j^\htmlClass{text-primary}{\vec{v}_{avg(2\to 4)} = \frac{\Delta\vec{r}_{2\to 4}}{\Delta t} = \frac{4\hat{i} - 12\hat{j}}{4-2} = 2\hat{i} - 6\hat{j}}
Instantaneous velocity v⃗(t)\vec{v}(t) or velocity vector at a specific instant. It is the time derivative of the position vector. Its direction is tangent to the trajectory.
v⃗(t)=dr⃗dt\vec{v}(t) = \frac{d\vec{r}}{dt}
v⃗(t)=2i^−2tj^\htmlClass{text-primary}{\vec{v}(t) = 2\hat{i} - 2t\hat{j}}
Speed: is the scalar value of velocity (magnitude).
∣v⃗(t)∣=vx2+vy2Ex.: ∣v⃗(t)∣=4+4t2|\vec{v}(t)| = \sqrt{v_x^2 + v_y^2} \quad \text{Ex.: } |\vec{v}(t)| = \sqrt{4+4t^2}
Velocity vector tangent to trajectory
Acceleration
Average acceleration, a⃗avg\vec{a}_{avg}: indicates the change in velocity between two instants with respect to the time elapsed.
a⃗avg=Δv⃗Δt=v⃗t2−v⃗t1t2−t1\vec{a}_{avg} = \frac{\Delta\vec{v}}{\Delta t} = \frac{\vec{v}_{t_2} - \vec{v}_{t_1}}{t_2 - t_1}
a⃗avg(2→4)=v⃗4−v⃗2Δt=−4j^4−2=−2j^\htmlClass{text-primary}{\vec{a}_{avg(2\to 4)} = \frac{\vec{v}_4 - \vec{v}_2}{\Delta t} = \frac{-4\hat{j}}{4-2} = -2\hat{j}}
Instantaneous acceleration a⃗(t)\vec{a}(t): acceleration vector at a specific instant. It is the time derivative of velocity.
a⃗(t)=dv⃗dt\vec{a}(t) = \frac{d\vec{v}}{dt}
a⃗(t)=−2j^\htmlClass{text-primary}{\vec{a}(t) = -2\hat{j}}
Intrinsic components of acceleration: a⃗\vec{a} can be decomposed into a component tangent to the trajectory, which causes the change in speed, and another component perpendicular to it, which causes the change in direction, satisfying: a2=at2+an2a^2 = a_t^2 + a_n^2
Intrinsic components of acceleration
Tangential acceleration: Its magnitude is:
∣a⃗t∣=∣d∣v⃗(t)∣dt∣|\vec{a}_t| = \left|\frac{d|\vec{v}(t)|}{dt}\right|
∣a⃗t∣=d(4+4t2)dt=4t4+4t2\htmlClass{text-primary}{|\vec{a}_t| = \frac{d(\sqrt{4+4t^2})}{dt} = \frac{4t}{\sqrt{4+4t^2}}}
Normal acceleration: or centripetal acceleration. It is directed towards the center of curvature. It allows calculating the radius of curvature. Its magnitude is:
∣a⃗n∣=∣v⃗∣2R|\vec{a}_n| = \frac{|\vec{v}|^2}{R}
Uniform Linear Motion (or Uniform Rectilinear Motion) is a motion with a constant velocity, v0v_0. The position equation is:
x=x0+v0⋅tx = x_0 + v_0 \cdot t
Uniform Linear Motion diagram
Uniformly Accelerated Linear Motion is a motion with a constant acceleration, aa.
Uniformly Accelerated Linear Motion diagram
The equations of motion are:
{x=x0+v0⋅t+12a⋅t2v=v0+a⋅t\begin{cases} x = x_0 + v_0 \cdot t + \frac{1}{2}a \cdot t^2 \\ v = v_0 + a \cdot t \end{cases}
Combining both yields:
v2=v02+2a(x−x0)v^2 = v_0^2 + 2a(x - x_0)
In an overtaking or meeting problem, we set up the appropriate equations of motion for each object;
x1=f(t)x2=g(t)x_1 = f(t) \quad x_2 = g(t)
The key is that when they meet, their positions are equal, so we can equate their equations and solve for the time they take to meet:
x1=x2  ⟹  f(t)=g(t)x_1 = x_2 \implies f(t) = g(t)
Example: A cyclist leaves the position x=1000 m with a constant velocity v0=5m/s. 10s later, a car starts from the origin to catch up with an acceleration a=2m/s². Calculate the time and position when it catches the cyclist.
To solve a kinematics problem, we will follow these steps:
1. Diagram and data: taking into account that the cyclist has a 10s head start, setting the time origin to when the car starts. We have this diagram:
Overtaking problem diagram
2. Set up the equations of motion for each object
Car UALM
Cyclist ULM
xcar=12⋅2⋅t2vcar=2⋅t\htmlClass{text-primary}{x_{car} = \frac{1}{2}\cdot 2 \cdot t^2 \quad v_{car} = 2 \cdot t}
xcyc=1000+5⋅(t+10)\htmlClass{text-primary}{x_{cyc} = 1000 + 5\cdot(t+10)}
3. Solve the equations: when they meet, their positions are equal, so equating xcyc=xcarx_{cyc} = x_{car}
1000+5⋅(t+10)=t2\htmlClass{text-primary}{1000 + 5\cdot(t+10) = t^2}
Of the two solutions, we discard the negative one. We get the time it takes the car to catch the cyclist. t=35s
Substituting into either position equation yields: x=1225m
This type of motion is a specific case of UALM (Uniformly Accelerated Linear Motion). If we set a=−ga=-g, and denote positions with the letter hh, the equations become:

{h=h0+v0⋅t−12g⋅t2v=v0−g⋅t\begin{cases} h = h_0 + v_0 \cdot t - \frac{1}{2}g \cdot t^2 \\ v = v_0 - g \cdot t \end{cases}
Vertical motion diagram
We \must consider some special cases:
• If the object is dropped from rest; v0=0v_0 = 0
• At the point of maxi\mum height; v=0v = 0
• Upon hitting the ground; h=0h = 0
A specific case is calculating the velocity with which a body dropped from a height hh hits the ground. It is obtained by combining both equations:
v=2ghv = \sqrt{2gh}
Vertical motion example

A ball is thrown upwards with an initial velocity of 15m/s, from a height of 25m. a) Calculate the maxi\mum height it will reach, and b) the velocity when it hits the ground.
1. Diagram:
2. Equations of motion:
Vertical motion example diagram
{h=25+15⋅t−4.9⋅t2v=15−9.8⋅t\htmlClass{text-primary}{\begin{cases} h = 25 + 15 \cdot t - 4.9 \cdot t^2 \\ v = 15 - 9.8 \cdot t \end{cases}}
3. a) At the highest point, v=0v=0, therefore
0=15−9.8⋅t  ⟹  t=1.53 s\htmlClass{text-primary}{0 = 15 - 9.8 \cdot t \implies t = 1.53\text{ s}}
Substituting into the position equation yields: h=36.48m
4. b) upon hitting the ground h=0h=0, then
0=25+15⋅t−4.9⋅t2\htmlClass{text-primary}{0 = 25 + 15 \cdot t - 4.9 \cdot t^2}
Solving this gives t=−1.19 st=-1.19\text{ s} and t=4.26 st=4.26\text{ s}. The negative value is discarded, and substituting t=4.26s into the velocity equation gives v=-26.74 m/s
Projectile motion is a composite motion along two axes where the object is launched with an initial velocity forming an angle α\alpha with the horizontal. The motion is broken down into its xx and yy components with the following characteristics:
General projectile motion diagram
Motion in X (ULM)
v0x=v0cos⁡α;ax=0v_{0x} = v_0 \cos \alpha; \quad a_x = 0
Equations:
x=x0+v0x⋅tx = x_0 + v_{0x} \cdot t
Motion in Y (UALM)
v0y=v0sin⁡α;ay=−gv_{0y} = v_0 \sin \alpha; \quad a_y = -g
Equations:
{y=y0+v0y⋅t−12g⋅t2vy=v0y−g⋅t\begin{cases} y = y_0 + v_{0y} \cdot t - \frac{1}{2}g \cdot t^2 \\ v_y = v_{0y} - g \cdot t \end{cases}
A ball is launched from the ground with v0=20m/s and α=30°. a) Calculate the maximum height and b) the maximum range
1. Diagram
2. Equations of motion:
Projectile motion example diagram
{x=20⋅cos⁡30∘⋅ty=20⋅sin⁡30∘⋅t−4.9⋅t2vy=20⋅sin⁡30∘−9.8⋅t\htmlClass{text-primary}{\begin{cases} x = 20 \cdot \cos 30^\circ \cdot t \\ y = 20 \cdot \sin 30^\circ \cdot t - 4.9 \cdot t^2 \\ v_y = 20 \cdot \sin 30^\circ - 9.8 \cdot t \end{cases}}
3. a) The maxi\mum height is reached when vy=0v_y=0, then
0=20⋅sin⁡30∘−9.8⋅t  ⟹  t=1.02 s\htmlClass{text-primary}{0 = 20 \cdot \sin 30^\circ - 9.8 \cdot t \implies t = 1.02\text{ s}}
Substituting into the vertical position equation yields: y=5.1 m
3. b) The maxi\mum range occurs when the ball hits the ground, that is, y=0y=0. From the yy-position equation we have:
0=20⋅sin⁡30∘⋅t−4.9⋅t2\htmlClass{text-primary}{0 = 20 \cdot \sin 30^\circ \cdot t - 4.9 \cdot t^2}
Solving this gives t=0t=0 and t=2.04 st=2.04\text{ s}. The first value is the moment of launch. We take the second value in the xx-equation to obtain xmax=35.33 m