Motion in Space

Motion in space involve the position \( \vec{r}(t) \), the velocity \( \vec{v}(t)=\vec{r'}(t) \), the acceleration \( \vec{a}(t)=\vec{v'}(t) \), and the speed \( v(t)=||\vec{v}(t)|| \) of an object.

The Essentials

The acceleration vector of a function can be split into two components, the tangential component and the normal component:

\[ \vec{a}(t)=a_T\vec{T}(t)+a_N\vec{N}(t) \]

Where the magnitude of the tangential acceleration is:

\[ a_T=\vec{a}\cdot\vec{T} \]

And the magnitude of the normal acceleration is:

\[ a_N=\vec{a}\cdot\vec{N} \]

Newton's second law in vector form is \( m\vec{a}=F_{net} \). In the case of projectile motion the only force is gravity assuming no air resistance \( F_{net}=-mg\hat{j} \). If we integrate both sides of the equation twice we end up with:

\[ \vec{s}(t)=-\frac{1}{2}gt^2\hat{j}+\vec{v}_0t+\vec{s}_0 \]

Where the initial velocity and position come from the constants of integration. If we set the initial position to be the origin \( \vec{s}=\vec{0} \), we can measure the motion of the projectile given an initial speed and the angle of projection:

\[ \vec{v}_0=v_0\cos(\theta)\hat{i}+v_0\sin(\theta)\hat{j} \]
\[ \vec{s}(t)=v_0t\cos(\theta)\hat{i}+(v_0t\sin(\theta)-\frac{1}{2}gt^2)\hat{j} \]

Example

A projectile is launched from 150 ft above the ground with a speed of 500 ft/sec at a 30 deg angle from horizontal. Find the maximum height of the projectile, how long it takes to hit the ground and how far it will travel horizontally.

First we will plug the values into the equation:

\[ \vec{s}(t)=250t\sqrt{3}\hat{i}+(250t-16t^2)\hat{j} \]

The maximum height is reached when the vertical component of the velocity is 0:

\[ v_y=250-32t=0 \]

The upward velocity is 0 when \( t=7.81 \). If we plug that into our vector position function, the vertical component of the position (the maximum height) at that time is: 977 ft.

If the launch happens at the origin, the projectile lands when the vertical component of position is -150 ft. We can write just the vertical components in the position equation and solve for t:

\[ -150=250t-16t^2 \]

Which is true when \( t=16.2 \).

By plugging 16.2 in for t in the equation and looking at the horizontal component of the position, we can find how far the projectile goes: 7010 ft.

Practice

  1. Find the velocity and acceleration of \( \vec{r}=\langle2\cos(t),2\sin(t),t^2\rangle \) at time \( t=\pi \)
  2. Find \( a_T \) and \( a_N \) for this function: \( \vec{r}=\langle2t^2,t^2-4,t+3\rangle \)

Solutions:

  1. \( \vec{v}(\pi)=\langle0,-2,2\pi\rangle \), \( \vec{a}(\pi)=\langle2,0,2\rangle \)
  2. \( a_T=\frac{20t}{\sqrt{20t^2+1}} \) \( a_N=\sqrt{\frac{20}{20t^2+1}} \)

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