Physics 09.02.2026

10060 - Physics/Kinematics 2D/Kinematics 2D Opgaver.typ

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+#import "@local/dtu-template:0.6.2":*
+
+#show: dtu-physics-note.with(
+    course: "10060",
+    author: "Rasmus Rosendahl-Kaa",
+    course-name: "Physics",
+    title: "Kinematics 2D - Opgaver",
+    date: datetime(year: 2026, month: 02, day: 09),
+    semester: "Spring 2026"
+)
+// Import Cetz for drawing diagrams
+#import "@preview/cetz:0.4.2"
+
+// Helper function for headers
+#let header(title) = {
+  v(1em)
+  text(weight: "bold", size: 1.2em, title)
+  line(length: 100%, stroke: 0.5pt)
+  v(0.5em)
+}
+
+#align(center)[
+  #text(size: 16pt, weight: "bold")[Chapter 4 - Problems]\
+  #text(size: 14pt)[Physics]
+]
+
+// --- PROBLEM 1 ---
+#header("Problem 1")
+
+A particle moves in a projectile motion neglecting air resistance. In the figure below, part of the parabolic trajectory is drawn, and three points, 1, 2 and 3, are marked. The speed of the particle in the points are $v_1$, $v_2$ and $v_3$.
+
+#align(center)[
+  #cetz.canvas({
+    import cetz.draw: *
+
+    // Draw Ground
+    line((-1, 0), (7, 0), stroke: (thickness: 2pt))
+
+    // Draw Parabola (Bezier)
+    // Start low left, peak middle high, end low right
+    bezier((0, 0), (4.5, 2.6), (2.7, 7.5), stroke: (dash: "dashed", paint: blue, thickness: 1.5pt), name: "curve")
+
+    // Mark points
+    let p1 = (0.75, 1.8)
+    let p2 = (3, 4.5) // Peak area
+    let p3 = (4.5, 2.6)
+
+    fill(black)
+    circle(p1, radius: 0.15, name: "c1")
+    circle(p2, radius: 0.15, name: "c2")
+    circle(p3, radius: 0.15, name: "c3")
+
+    content((p1.at(0) + 0.4, p1.at(1)), text(size: 14pt, "1"))
+    content((p2.at(0), p2.at(1) - 0.5), text(size: 14pt, "2"))
+    content((p3.at(0) - 0.4, p3.at(1)), text(size: 14pt, "3"))
+
+    // Draw Direction Arrows on path
+    line((1.5, 3.5), (1.8, 4.0), mark: (end: "stealth", fill: blue, scale: 0.8), stroke: (paint: blue, thickness: 2pt))
+    line((4.5, 4.0), (4.8, 3.5), mark: (end: "stealth", fill: blue, scale: 0.8), stroke: (paint: blue, thickness: 2pt))
+  })
+]
+
+What is the relationship between the speed in the three points?
+
+#grid(
+  columns: (1fr, 1fr),
+  gutter: 1em,
+  [A) $v_1 = v_2 = v_3$],
+  [B) $v_1 < v_2 < v_3$],
+  [C) $v_1 < v_3 < v_2$],
+  [D) $v_2 < v_1 < v_3$],
+  [E) $v_2 < v_3 < v_1$],
+  [F) $v_3 < v_1 < v_2$],
+  [G) $v_3 < v_1 < v_2$],
+  [H) Don't know]
+)
+
+#solution()[
+*E)*
+
+Da der ingen luftmodstand er kan vi se bort fra den da den må være ens i alle punkter.
+
+$v_2$ er mindst fordi der står den stille lodret.
+
+$v_3$ er mindre end $v_1$ fordi den er højere oppe. Den tid/afstand det tager for en hastighed at blive 0 mod konstant acceleration, må hastigheden være den samme i modsat retning samme sted men på vej ned.
+
+]
+
+// --- PROBLEM 2 ---
+#header("Problem 2")
+
+A person kicks a ball which rests on the top of a staircase. The ball has a horizontal velocity $v$ after the kick. The ball just misses the lowest step on the staircase. Each step on the staircase is a square with side lengths $d$.
+
+#align(center)[
+  #cetz.canvas({
+    import cetz.draw: *
+
+    // Draw Stairs
+    let step_size = 1
+    let steps = 3
+
+    // Top platform
+    line((-2, 3), (0, 3), stroke: 1pt)
+
+    // Steps
+    line((0,3), (0,2), (1,2), (1,1), (2,1), (2,0), (3,0), (3,-1), stroke: 1pt)
+
+    // Ground
+    line((-3,-1), (4,-1), stroke: 2pt)
+
+    // Ball
+    circle((0, 3.15), radius: 0.15, fill: black)
+    line((0, 3.15), (1, 3.15), mark: (end: "stealth"), name: "vec")
+    content("vec.end", anchor: "south", padding: 0.2, $v$)
+
+    // Trajectory
+    bezier((1, 3.15), (3, 0), (2.1, 3), stroke: (dash: "dashed"))
+
+    // Dimensions
+    line((-0.5, 2), (-0.5, 1), mark: (start: "stealth", end: "stealth"))
+    content((-0.8, 1.5), $d$)
+
+    line((1, 0.5), (2, 0.5), mark: (start: "stealth", end: "stealth"))
+    content((1.5, 0.8), $d$)
+  })
+]
+
+What is the magnitude of the initial velocity $v$?
+
+#grid(
+  columns: (1fr, 1fr, 1fr),
+  row-gutter: 1em,
+  [A) $v = frac(sqrt(g d), 2)$],
+  [B) $v = sqrt(g d)$],
+  [C) $v = sqrt(3g d)$],
+  [D) $v = 3sqrt(g d)$],
+  [E) $v = frac(sqrt(3g d), sqrt(2))$],
+  [F) $v = frac(sqrt(3g d), 2)$],
+  [G) $v = frac(3sqrt(g d), sqrt(2))$],
+  [H) $v = frac(sqrt(3g d), 4)$],
+  [I) Don't know]
+)
+
+#solution()[
+Vi ved: $3 d = v dot t$
+
+Da den må have bevæget sig lige langt i x- og y-retning til sidst, må $v$ være gennemsnitshastigheden, bolden bevæger sig i, i y-retningen, aka $v_"slut"/2$:
+
+$
+v_"slut"^2 = 0 + 2 dot g (3 d)\
+v_"slut"^2 = 2 dot 3 g d\
+v_"slut" = sqrt(6 g d)
+$
+
+Svar er *C)* *forkert*
+]
+
+// --- PROBLEM 3 ---
+#header("Problem 3")
+
+A person is at equator and is moving in a circular motion. The radius of the earth is 6370 km.
+
+/ a): Determine the acceleration of the person.
+
+If earth were rotating faster, the person would become weightless when the acceleration was $g$.
+
+/ b): Determine the period of rotation of the earth if a person had to become weightless.
+
+#solution()[
+/ a):
+It takes 24 hours for person to make one rotation
+$
+a_"rad" = (4 pi^2 r)/T^2 = (4 pi^2 dot 6370 k m)/((24 h)^2) approx 0,0337 m/s^2
+$
+/ b):
+Now we have the $a_"rad"$ so use same formular as before, but isolate for $T$
+$
+9.82 m/s^2 = (4 pi^2 6370 k m)/(T^2) <=>\
+T^2 = (4 pi^2 6370 k m)/(9.82 m/s^2)\
+T^2 = (4 pi^2 dot (6370 dot 10^3) m)/(9.82 m/s^2)\
+T = sqrt((4 pi^2 dot (6370 dot 10^3) m)/(9.82 m/s^2)) approx 5060 s\
+$
+5060 seconds is about 1 hour and 24 minutes or 1.4 hours
+]
+
+// --- PROBLEM 4 ---
+#header("Problem 4")
+
+A rock is thrown from a point A on top of a tower and hits the point B on the ground 3.5 s later. The line from A to B has an angle of $50^(degree)$ with the horizon.
+
+#align(center)[
+  #cetz.canvas({
+    import cetz.draw: *
+
+    // Tower
+    rect((-1, 0), (0, 4), fill: pattern(size: (5pt, 5pt))[ \ \ / \ ], stroke: 1pt)
+    line((-1, 0), (6, 0), stroke: 1pt) // Ground
+
+    let start = (0, 4)
+    let end = (5, 0)
+
+    // Trajectory
+    bezier(start, end, (4, 4), stroke: (dash: "dashed"))
+
+    // Line A to B
+    line(start, end, stroke: (dash: "dashed"))
+
+    // Horizontal line from A
+    line(start, (2, 4), stroke: 0.5pt)
+
+    // Angle arc
+    arc((0.5,4), start: 0deg, stop: -50deg, radius: 0.4, name: "angle")
+    content((1.3, 3.6), $50^(degree)$)
+
+    // Points
+    content(start, anchor: "south-east", [A])
+    content(end, anchor: "north-west", [B])
+
+    // Initial velocity arrow (approx)
+    line(start, (2, 4), mark: (end: "stealth"))
+  })
+]
+
+a) Determine the initial speed of the rock.
+
+#solution()[
+Start height: $1/2 dot 9.82 dot 3.5^2 = 60.1475 m$
+
+Calculate distance to B:
+$
+tan(40^degree) = B/60.1475\
+B = tan(40^degree) dot 60.1475 approx 50.47 m
+$
+
+Hastighed må være $B/t = 50.47/3.5 = 14.42 m/s$
+]
+
+// --- PROBLEM 5 ---
+#header("Problem 5")
+
+A person plays with a tennis ball in a large room and tries to throw the ball from one end of the room and hit a picture hanging $h=30$ cm from the ceiling on the opposite end of the room. It is assumed, that the person throwing the ball, throws the ball in the same height as the picture. The ball just barely avoids touching the ceiling. The room is $L=10$ m long.
+
+a) Determine the initial speed of the ball, the initial angle and duration of flight if the ball has to hit the picture.
+
+#solution()[
+We know the following things:
+$
+v_y (t^*) = 0\
+y(t^*) = 30 c m\
+x(T/2) = 5 m\
+$
+To get initial angle:
+$
+theta &= arctan((2 dot 2 dot h)/L)\
+&= arctan((2 dot 2 dot 30)/(10 dot 100)) = 3/25 "rad" = 6.875^degree
+$
+Initial speed:
+$
+v_0 = sqrt((h dot 2g)/sin^2 (theta)) = sqrt((0.3m dot 2 dot 9.82 m/s^2)/sin^2 (3/25)) approx 20.28 m/s approx 73 "km"/t
+$
+Time:
+$
+T_"tot" = (2 dot v_0 dot sin(theta))/g = (2 dot 20.28 dot sin(3/25))/9.82 approx 0.5 s
+$
+
+]
+
+// --- PROBLEM 6 ---
+#header("Problem 6")
+
+A diver jumps from 3 m springboard. During the jump, the diver reaches a height on 2.5 m above the springboard and lands in the water 2.8 meters from the starting point.
+
+a) Determine the x and y components of the initial speed, the angle of the jump as well as the duration of flight.
+
+#solution()[
+*Initial y-speed:*
+$
+t &= - v_0/(-g) = v_0/g\
+2.5 m &= v_0 dot v_0/g + 1/2 g (v_0/g)^2\
+&= v_0^2/g + 1/2 g (v_0^2)/g^2\
+&= v_0^2/g + 1/2 (v_0^2)/g\
+&=(3 v_0^2)/(2 g)\
+3v_0^2 &= 2.5 m dot 2 g\
+v_0 &approx 7 m/s \
+$
+
+*Total time:*
+$
+0 = 3 m + 7 m/s dot t s - 1/2 dot 9.82 m/s^2 dot t^2\
+t approx 1.77 s
+$
+
+*Inital x-speed:*
+$
+2.8 m = v_(0x) dot 1.77\
+v_(0x) = 2.8/1.77 approx 1.58 m/s
+$
+
+*Angle:*
+
+To find angle, we can use trigonometry. We can make the triangle for when the diver reaches top height after $7/9.82$ seconds.
+
+Height is 2.5, length is $1.58 dot 7/9.82 approx 1.13 m$
+
+Angle must be $arctan(2.5/1.13) approx 1.147 "rad" approx 65.72^deg$
+
+]
+
+// --- PROBLEM 7 ---
+#header("Problem 7")
+
+In a shopping cart a vertical toy cannon is placed. The shopping cart is traveling with constant speed $v$ on the floor in the supermarket. The cannon shots by mistake a projectile vertical into the air compared to the shopping cart, with the same velocity $v$ as the shopping cart is traveling with. By accident, the projectile from the cannon end up into the cannon after the shopping cart has traveled the distance $L$.
+
+#align(center)[
+  #cetz.canvas({
+    import cetz.draw: *
+
+    let cart(x) = {
+      rect((x, 0.5), (x + 1.5, 1.5), stroke: 1.5pt) // Basket
+      line((x, 1.5), (x - 0.3, 2), stroke: 1.5pt) // Handle
+      circle((x + 0.3, 0.25), radius: 0.25, stroke: 1.5pt) // Wheel 1
+      circle((x + 1.2, 0.25), radius: 0.25, stroke: 1.5pt) // Wheel 2
+      rect((x + 0.5, 0.5), (x + 0.7, 1.8), fill: blue) // Cannon
+    }
+
+    // Floor
+    rect((-1, -0.5), (8, 0), fill: black)
+
+    // Cart 1
+    cart(0)
+    line((1.6, 1), (2.5, 1), mark: (end: "stealth"))
+    content((2, 1.2), $v$)
+
+    // Cart 2
+    cart(5)
+
+    // Trajectory
+    bezier((0.6, 1.8), (2.5, 5), (5.6, 1.8), stroke: (dash: "dashed"))
+
+    // Distance L
+    line((0.75, -0.8), (5.75, -0.8), mark: (start: "stealth", end: "stealth"))
+    content((3.25, -1.2), $L$)
+  })
+]
+
+What is the velocity $v$ which makes this maneuver possible?
+
+#grid(
+  columns: (1fr, 1fr, 1fr),
+  row-gutter: 1em,
+  [A) $v = sqrt(g L)$],
+  [B) $v = sqrt(2g L)$],
+  [C) $v = sqrt(2sqrt(2) g L)$],
+  [D) $v = sqrt((g L)/2)$],
+  [E) $v = sqrt((g L)/(2sqrt(2)))$],
+  [F) Don't know]
+)
+
+// --- PROBLEM 8 ---
+#header("Problem 8")
+
+A ball is thrown from one person to another, with a distance $L$ between one another. The ball is thrown and catched at the same height. The ball is thrown in a way, so the ball is at the receiver with lowest speed possible.
+
+a) What is the speed of the ball immediately before it is catched?
+b) At what angle should the ball be thrown?
+
+// --- PROBLEM 9 ---
+#header("Problem 9")
+
+A particle (1) is at time $t=0$ thrown with an initial velocity $v_0$ from an unknown height $h$ above the ground. At the same time, another particle (2) with initial velocity is thrown along the ground. Particle (2) starts its horizontal movement at point $L$ to the right of the first particle. Neglect any friction. At an unknown time $T$, the two particles collide.
+
+#align(center)[
+  #cetz.canvas({
+    import cetz.draw: *
+
+    // Cliff
+    line((-1, 4), (0, 4), (0, 0), (6, 0), stroke: 1.5pt)
+
+    // Particle 1
+    circle((0, 4), radius: 0.1, fill: black)
+    content((0, 4), anchor: "south-east", [(1)])
+    line((0, 4), (1.5, 4), mark: (end: "stealth"))
+    content((0.75, 4.3), $v_0$)
+
+    // Particle 2
+    circle((3, 0), radius: 0.1, fill: black)
+    content((3, 0), anchor: "south-east", [(2)])
+    line((3, 0), (4.5, 0), mark: (end: "stealth"))
+    content((3.75, 0.3), $v_0/2$)
+
+    // Trajectory
+    bezier((0, 4), (3, 4), (5.5, 0), stroke: (dash: "dashed"))
+
+    // Dimensions
+    line((-0.2, 4), (-0.2, 0), mark: (start: "stealth", end: "stealth"))
+    content((-0.5, 2), $h$)
+
+    line((0, -0.5), (3, -0.5), mark: (start: "stealth", end: "stealth"))
+    content((1.5, -0.8), $L$)
+  })
+]
+
+a) Determine the time $T$.
+b) Determine the height $h$.
+
+The height is assumed known:
+c) What is the magnitude of the relative speed, $vec(v)_(r e l) = vec(v)_1 - vec(v)_2$ between the two particles immediately before they collide?
+
+// --- PROBLEM 10 ---
+#header("Problem 10")
+
+A football player has a free kick $3L = 30.0$ m from the goal. The wall of defenders is placed 10.0 m from where the free kick is taken. The wall of defenders is $h=2.0$ m tall. The goal has the height $H=2.44$ m. The football player kicks the ball, so the initial velocity has an angle of $15^(degree)$ with the horizon, and he kicks it hard enough, so the ball does not hit the ground before it hits the goal.
+
+a) Make a drawing of the situation with relevant physical magnitudes.
+b) Determine the initial velocity the football player has to kick the ball, so the ball just barely goes over the wall.
+c) Determine the interval of initial speed in which the ball ends in the goal.
+
+// --- PROBLEM 11 ---
+#header("Problem 11")
+
+A car travels in on a circular track with radius $R$. The car starts from rest and has a tangential acceleration $a_0$.
+
+a) Derive an expression for the speed of the car and the radial acceleration.
+b) Determine the magnitude of the acceleration of the car when it has traveled one round on the track.

10060 - Physics/Kinematics 2D/Kinematics 2D.typ

@@ -0,0 +1,264 @@
+#import "@local/dtu-template:0.6.2":*
+
+#show: dtu-physics-note.with(
+    course: "10060",
+    author: "Rasmus Rosendahl-Kaa",
+    course-name: "Physics",
+    title: "Kinematics 2D",
+    date: datetime(year: 2026, month: 02, day: 09),
+    semester: "Spring 2026"
+)
+
+= Hastighedsvektor i flere dimensioner
+Laver en positionsvektor
+$
+arrow(r)(t)=vec(x(t), y(t), z(t))
+$
+
+For at finde ændring i position:
+$
+Delta r = r_2 - r_1
+$
+
+Gennemsnitlig hastighedsvektor:
+$
+arrow(v)_("av") = (Delta r)/(Delta t) = (r_2 - r_1)/(t_2 - t_1)
+$
+
+== Instantan hastighedsvektor
+$
+arrow(v) = vec( (d x(t))/( d t), (d y(t))/( d t), (d z(t))/( d t) ) = vec(v_x, v_y, v_z) = vec(x'(t), y'(t), z'(t))
+$
+
+Fart:
+$
+|arrow(v)| = sqrt(v_x^2 + v_y^2 + v_z^2)
+$
+
+== Komponenter af hastighedsvektor
+Hvis du har vinkel for hastighedsvektor og hastighedsvektoren, så kan du få $v_x (t), v_y (t)$:
+$
+v_x = v(t) dot cos(alpha)\
+v_y = v(t) dot sin(alpha)
+$
+Fart:
+$
+|arrow(v)| = sqrt(v_x^2 + v_y^2)
+$
+Finde vinkel/retning:
+$
+tan(alpha) = v_y/v_x\
+alpha = arctan(v_y/v_x)
+$
+
+= Accelerationsvektor
+Er en ændring i hastighedsvektor
+$
+Delta arrow(v) = arrow(v_2) - arrow(v_1)
+$
+Gennemsnitlig acceleration
+$
+arrow(a)_("av") = (Delta arrow(v))/(Delta t)
+$
+
+== Instantan accelerationsvektor
+Når $Delta t -> 0$:
+
+$
+arrow(a)=(d arrow(v))/(d t) = vec((d v_x)/(d t), (d v_y)/(d t), (d v_z)/(d t)) = vec(a_x, a_y, a_z)
+$
+
+== Parallel og vinkelret komponent af accelerationsvektor
+Kan finde komponenterne af accelerationsvektor
+
+Tangentiel $a_||$: ændring i $|arrow(v)|$
+
+Radiel $a_perp$: ændring af $arrow(v)$
+
+= Projektilbevægelse
+Sker kun i to retninger: x og y.
+
+Er nogle forudsætninger:
+/ 1): Kun tyngdekraft. Dvs. ingen luftmodstand
+/ 2): Punktformet partikel
+
+Vil finde $(x(t), y(t))$
+
+#note-box(title: "Fra forlæsning 1")[
+Formler for bevægelse med konstant acceleration i én dimension:
+$
+v = v_0 + a t\
+x = x_0 + v_0 t + 1/2 a t^2
+$
+]
+
+*Acceleration:* (er konstant i tid):
+$
+a_x = 0\
+a_y = -g\
+$
+
+*Hastighed:*
+$
+v_x = v_(0x) = |arrow(v)_0| dot cos(alpha_0)\
+v_y = v_(0y) - g dot t = |arrow(v)_0| dot sin(alpha_0) - g dot t
+$
+
+*Position:*
+$
+x(t)=x_0 + v_(0x) dot t "(er lineær i tid)"\
+y(t)=y_0 + v_(0y) dot t - 1/2 g t^2 "(er kvadratisk i tid)"
+$
+
+== Banekurve
+Vil gerne finde $y(x)$ (y, som er afhængig af x).\
+Antager: $x_0 = y_0 = 0$
+
+$
+x(t) = v_0 cos(alpha_0) dot t\
+y(t) = v_0 sin(alpha_0) dot t - 1/2 g dot t^2
+$
+
+Isolér $t$ i $x(t)$ og sæt ind i $y(t)$:
+$
+t &= x/(v_0 cos(alpha_0))\
+y(t) &= (v_0 sin(alpha_0) dot x)/(v_0 cos(alpha_0)) - 1/2 g dot (x/(v_0 cos(alpha_0)))^2\
+&=tan(alpha_0) dot x - g/(2 v_0^2 cos^2 (alpha_0)) dot x^2
+$
+
+Der *skal* startes i 0 før ligningen gælder, men der behøves ikke at ende i 0.
+
+#example()[
+2 lande er i krig: Det ene land beskyder det andet land med et missil. Missilet skal rejse $L=30 "km"$ for at ramme målet og holde sig under $h=1 "km"$ højde for ikke at blive skudt ned.
+
+/ a): Hvad er hastigheden på missilet? Og hvad er affyringsvinklen?
+/ b): Hvor lang tid er missilet undervejs?
+
+#solution()[
+Højeste punkt:
+$
+h = y = (v_0 dot sin(theta))^2/(2 g) <=> v_0 = sqrt((h dot 2 g)/(sin^2 (theta)))
+$
+Længde:
+$
+L = x = (v_0^2 dot sin(2 theta))/g = (h 2 g sin(2 theta))/(g sin^2 (theta)) = (2 dot 2 h cos(theta))/(sin(theta))\
+<=> (sin(theta))/(cos(theta)) = (2 h)/L <=> tan(theta) = (2 dot 2 h)/L\
+theta = arctan((2 dot 2 h)/L) = arctan((4 dot 1000 m)/(30 dot 1000 m)) = 0.13 "rad" approx 7.59^degree
+$
+$
+v_0 = sqrt((h dot 2 g)/(sin^2 (theta))) = sqrt((1000 m dot 2 dot 9.8 m/s^2)/(sin^2 (0.13 "rad"))) = 1059 m/s approx 3813 (k m)/t
+$
+
+$
+T_("tot") = (2 dot v_0 dot sin(theta))/g = (2 dot 1059 m/s dot sin(0.13 "rad"))/(9.8 m/s^2) = 28 s
+$
+]
+]
+
+
+#example()[
+2 lande er I krig: Det ene land beskyder det andet land fra et bjerg på 500 m med et missil. Missilet skal rejse
+L=30 km for at ramme målet og holde sig under h=1 km højde for ikke at blive skudt ned.
+
+/ a): Hvad er hastigheden på missile ? Og hvad er affyrings vinklen?
+/ b): Hvor lang tid er missile undervejs?
+
+#solution()[
+Opskriv ligninger
+$
+x(t)=v_(0x) dot t\
+y(t) = (v_0y) dot t - 1/2 g t^2 + 500 m\
+v_x = v_(0x)\
+v_y = v_(0y) - g t
+$
+
+Lad T være tidspunktet for impakt:
+
+$
+x(t = T) = L = v_(0x) dot T\
+y(t = T) = 0 = v_(0y) dot T - 1/2 g T^2
+$
+
+Lad $t^*$ være når missilet når toppunktet:
+$
+v_x (t^*) = v_(0x)\
+v_y (t^*) = 0 = v_(0y) dot t^*
+$
+
+$
+v_(0y)/v_(0x) = (v_0 sin(theta))/(v_0 cos(theta)) <=> tan(theta) = v_(0y)/v_(0x)
+$
+
+Har 5 ligninger med 5 ubekendte. Brug Python/Maple til at udregne og finde de ubekendte.
+]
+]
+
+
+= Cirkulær bevægelse med konstant fart
+Dvs. $|arrow(v)|$ er konstant. Accelerationen vender indad mod orego.
+
+$|arrow(a)_("rad")| = v^2/r$
+
+$v = ("omkreds")/("omløbstid") = (2 pi r)/T$
+
+$
+a_("rad") = ((2 pi r)/T)^2/r = (4 pi^2 dot r^2)/(T^2 dot r) = (4 pi^2 dot r)/T^2
+$
+
+= Cirkulær bevægelse med ikke-konstant fart
+$
+a_("rad") = v^2/r\
+a_("tan") = (d dot |v|)/(d t)
+$
+
+Obs $(d |v|)/(d t) eq.not |(d v)/(d t)|$. Det er forskellen i fart, ikke forskellen i hastighed.
+
+
+$
+(d arrow(v))/(d t) = arrow(a)_("total")\
+|(d v)/(d t)| = |a_("total")|
+$
+
+#example()[
+En centrifuge har en konstant tangential acceleration på $a_t = 8 m/s^2$ i 15 s hvorefter den opnår centrifuge hastigheden.
+
+/ a): Hvad er denne hastighed?
+/ b): Hvor mange omdrejninger per minut (rpm) har centrifugen efter accelerationen?
+
+#solution()[
+/ a):
+$(d |v|)/(d t) = a_t$
+$
+v(t) &= integral a_t d t = integral_0^(15 s) a_t d t = [a_t dot t]_0^(15 s)\
+&= 8 m/s^2 dot 15 s = 120 m/s
+$
+/ b):
+Find omkreds:
+$
+0 = 2 pi dot r = 2 pi dot 0.3 m = 1.8 m\
+"rpm" = (120 m/s)/(1.8 m) = 56.5 "rps" dot 60 = 3400 "rpm"
+$
+
+]
+]
+
+= Relativ hastighed i to dimensioner
+
+#image("Billeder/Relativ-hastighed.png", width: 50%)
+
+$
+arrow(v)_(F|J) = arrow(v)_(F|L) + arrow(v)_(L|J)\
+
+$
+
+
+#example()[
+#image("Billeder/Relativ-hastighed-eksempel.png")
+
+#solution()[
+$
+sin(beta) = "Modstående katete"/"Hypotenusen"\
+= |v_(L|J)|
+$
+]
+]