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Wize University Physics Textbook (Master) > Motion in 1/2/3D: Kinematics

Two- and Three-Dimensional Kinematics (Calculus-Based)

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Two- and Three-Dimensional Kinematics: Part 1


  • In general, the position of a particle is represented by a 3D vector that is changing with time. The components of this 3D vector are a function of time too.
r(t)⟶=x(t)ı^+y(t)ȷ^+z(t)k^\boxed{\stackrel{\longrightarrow}{r(t)}=x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat k}







  • Displacement is a vector that shows the change in position of the particle. The unit of displacement is m.

Δr⟶=rf→−rı^→=Δxı^+Δyȷ^+Δzk^\stackrel{\longrightarrow}{\Delta r}=\stackrel{\rightarrow}{r_f}-\stackrel{\rightarrow}{r_{\hat{\imath}}}=\Delta x\hat{\imath}+\Delta y\hat{\jmath}+\Delta z\hat k


  • Distance is the length of the path travelled by the particle



  • Average velocity is defined as

vave⟶=Δr⟶Δt\boxed{\stackrel{\longrightarrow}{v_{ave}}=\frac{\stackrel{\longrightarrow}{\Delta r}}{{\Delta t}}}






  • Instantaneous velocity is the velocity of an object in a specific time or position which in general might be different from the average velocity
v(t)⟶=vxı^+vyȷ^+vzk^=lim⁡Δt→0Δr⟶Δt=dr(t)⟶dt\stackrel{\longrightarrow}{v(t)}=v_x\hat{\imath}+v_y\hat{\jmath}+v_z\hat k=\lim_{\Delta t\to0}\frac{\stackrel{\longrightarrow}{\Delta r}}{{\Delta t}}=\frac{\stackrel{\longrightarrow}{dr(t)}}{{dt}}
dr(t)⟶dt=dx(t)dtı^+dy(t)dtȷ^+dz(t)dtk^\frac{\stackrel{\longrightarrow}{dr(t)}}{{dt}}=\frac{dx(t)}{dt}\hat{\imath}+\frac{dy(t)}{dt}\hat{\jmath}+\frac{dz(t)}{dt}\hat k
  • So, basically, the instantaneous velocity is obtained by taking the derivative of position with respect to time.
Wize Tip
Average speed and instantaneous speed could be found similarly if displacement vector is replaced by distance in above equations.



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Two- and Three-Dimensional Kinematics : Part 2


  • Instantaneous velocity is the velocity of an object in a specific time or position which in general might be different from the average velocity
v(t)⟶=vxı^+vyȷ^+vzk^=lim⁡Δt→0Δr⟶Δt=dr(t)⟶dt\stackrel{\longrightarrow}{v(t)}=v_x\hat{\imath}+v_y\hat{\jmath}+v_z\hat k=\lim_{\Delta t\to0}\frac{\stackrel{\longrightarrow}{\Delta r}}{{\Delta t}}=\frac{\stackrel{\longrightarrow}{dr(t)}}{{dt}}
dr(t)⟶dt=dx(t)dtı^+dy(t)dtȷ^+dz(t)dtk^\frac{\stackrel{\longrightarrow}{dr(t)}}{{dt}}=\frac{dx(t)}{dt}\hat{\imath}+\frac{dy(t)}{dt}\hat{\jmath}+\frac{dz(t)}{dt}\hat k
  • So, basically, the instantaneous velocity is obtained by taking the derivative of position with respect to time.
Wize Tip
Average speed and instantaneous speed could be found similarly if displacement vector is replaced by distance in above equations.

  • The displacement can be calculated by integrating the velocity vector over time.
Δr⟶=r⃗f−r⃗i=∫titf(vx(t)ı^+vy(t)ȷ^+vz(t)k^)dt\stackrel{\longrightarrow}{\Delta r}=\vec r_f-\vec r_i=\int^{t_f}_{t_i}(v_x(t)\hat{\imath}+v_y(t)\hat{\jmath}+v_z(t)\hat k)dt







  • Similarly, we have following relations between velocity and average acceleration:
aave⟶=ax(t)ı^+ay(t)ȷ^+az(t)k^=Δv⟶Δt\stackrel{\longrightarrow}{a_{ave}}=a_x(t)\hat{\imath}+a_y(t)\hat{\jmath}+a_z(t)\hat k=\frac{\stackrel{\longrightarrow}{\Delta v}}{{\Delta t}}

  • The instantaneous acceleration is the derivative of velocity with respect to time.
a(t)⟶=lim⁡Δt→0Δv⟶Δt=dv(t)⟶dt\stackrel{\longrightarrow}{a(t)}=\lim_{\Delta t\to0}\frac{\stackrel{\longrightarrow}{\Delta v}}{{\Delta t}}=\frac{\stackrel{\longrightarrow}{dv(t)}}{dt}
dv(t)⟶dt=dvx(t)dtı^+dvy(t)dtȷ^+dvz(t)dtk^\frac{\stackrel{\longrightarrow}{dv(t)}}{dt}=\frac{dv_x(t)}{dt}\hat{\imath}+\frac{dv_y(t)}{dt}\hat{\jmath}+\frac{dv_z(t)}{dt}\hat k

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  • Finally, we have
Δv⟶=v⃗f−v⃗i=∫titf(ax(t)ı^+ay(t)ȷ^+az(t)k^)dt\stackrel{\longrightarrow}{\Delta v}=\vec v_f-\vec v_i=\int^{t_f}_{t_i}(a_x(t)\hat{\imath}+a_y(t)\hat{\jmath}+a_z(t)\hat {k})dt
Wize Concept
Going from x⃗→v⃗→a⃗\vec{x} \rightarrow \vec{v} \rightarrow \vec{a} , you are taking the derivative (looking at the slope).
Going from x⃗←v⃗←a⃗\vec{x} \leftarrow \vec{v} \leftarrow \vec{a} , you are taking the integral (looking at the total area under the curve).




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Example: Velocity and Acceleration in 3D


The position of a particle in 3D is given by
r⃗(t)=(10t+t2+7)i^+2j^+4tk^  m\vec{r}(t)=(10t+t^2+7)\hat{i}+2\hat{j}+4t\hat{k} \ \ m
a) Calculate instantaneous velocity at t =5 sec
b) Calculate the average velocity for an interval (3-5 sec).
c) Calculate the instantaneous acceleration at t =1 sec.

Solution:

Part a)

We know that Instantaneous velocity is
v⃗=dr⃗dt\vec{v} = \dfrac{d\vec{r}}{dt}

dr⃗dt=(10+2t)i^+0j^+4k^\frac{d\vec{r}}{dt}=(10+2t)\hat{i}+0\hat{j}+4\hat{k}

v⃗(t=5)=(10+2(5))i^+4k^=20i^+4k^ m/s\vec{v}(t=5)= (10 + 2 (5))\hat{i}+4\hat{k} = 20\hat{i}+4\hat{k} \ m/s

Part b)

Average velocity:

v⃗average=Δr⃗Δt=r⃗(5)−r⃗(3)5−3\vec{v}_{average}=\dfrac{\Delta\vec{r}}{\Delta t}=\dfrac{\vec{r}(5)-\vec{r}(3)}{5-3}

v⃗average=[82i^+2j^+20k^]−[46i^+2j^+12k^]2=36i^+8k^2=18i^+4k^ m/s\vec{v}_{average} = \dfrac{[82\hat{i}+2\hat{j}+20\hat{k}]-[46\hat{i}+2\hat{j}+12\hat{k}]}{2}=\dfrac{36\hat{i}+8\hat{k}}{2}= 18\hat{i}+4\hat{k}\ m/s

Part c)

Instantaneous acceleration:
a⃗instant=d2r⃗dt2=2i^ ms2\vec{a}_{instan t}=\dfrac{d^2\vec{r}}{dt^2}=2\hat{i} \ \dfrac{m}{s^2}




A man stands on the roof of a 15.0-m-tall building and throws a rock with a velocity of magnitude 30.0 m/s at an angle of 33.0∘33.0^{\circ} above the horizontal. You can ignore air resistance.
Calculate
(a) the maximum height above the roof reached by rock
(b) the magnitude of the y-component of velocity of the rock just before it strikes the ground
(c) the horizontal range from the base of building to the point where the rock strikes the ground.
(d) Draw x−t,y−t,vxt, and vy−t x-t,y-t,v_xt,\ and\ v_y-t\ graphs for the motion.



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(a) the maximum height above the roof reached by rock

Extra Practice
Practice: T-shirt cannon
At football games, sometimes, there is a "T-shirt cannon" that is used to fire T-shirts into the audience, to be given to avid fans as souvenirs. In this problem, assume the flying T-shirts encounter no air resistance, and move like normal projectiles.
A T-shirt is launched with an initial velocity of v⃗=2.8msi^+13.3msj^\vec v=2.8 \frac {m}{s}\hat i+13.3\frac {m}{s}\hat j.
(Choose the direction of motion as positive; upward for y-direction, and in the direction of the T-shirt's motion for x-direction.)
Archery
You are hanging out with a friend at an archery range. Your friend is a lot stronger than you are, so her shots always have higher initial speed.
You should include rough sketches in your explanations for each of the following questions. It makes it easier for both you and the marker during a final exam.
a) You both shoot arrows horizontally at the same time from the same height. In what order do the arrows hit the ground?
A bird is swooping down at 23.0o below the horizontal at a speed of 12.3m/s. It spots a man, 25.0m in front, having his lunch below. How high up from the ground should the bird release in order to ruin the man's day? (answer in metres)
2D kinematics - meatball
You sneezed and your meatball rolls of the 0.60m tall table with a horizontal velocity of 2.4m/s.