Mv3–Mv5 - Equation of a particle's path

Mv3–Mv5 - Equation of a particle's path

Eliminate time to describe where a particle moves. You will retain the physical section of a curve, derive a projectile trajectory, and use a quadratic in the launch angle to decide whether a target is reachable.

Eliminating time without losing the domain

A particle has x=2tx=2t, y=3t2y=3t^2, for 0≤t≤30\le t\le3. Does it travel along an entire parabola?

No: the time interval limits its horizontal coordinate. Eliminating a parameter gives a curve containing the motion; permitted parameter values select the part actually travelled.

You already know

A position vector supplies one coordinate expression per component. Solve a convenient coordinate for the parameter and substitute into the other. Retain any physical time restrictions.

A Cartesian equation describes the route, but generally does not say how quickly the particle follows it.

Worked example 1: a finite parabolic arc

Question. Eliminate tt from x=2tx=2t, y=3t2y=3t^2, where 0≤t≤30\le t\le3, and identify the travelled path.

Plan. Solve the linear coordinate for time, substitute and translate the bounds.

t=x2first coordinatey=3(x2)2=34x2substitute0≤t≤3  ⟹  0≤x≤6retain bounds\begin{aligned} t&=\frac x2 && \text{first coordinate}\\ y&=3\left(\frac x2\right)^2=\frac34x^2 && \text{substitute}\\ 0\le t\le3&\implies0\le x\le6 && \text{retain bounds} \end{aligned}

Why this step? The equation alone includes negative xx, which this particle never reaches.

Answer. y=34x2y=\frac34x^2, for 0≤x≤60\le x\le6.

Check. At t=3t=3, the endpoint is (6,27)(6,27), on the stated curve.

Change only the initial horizontal position: xx now starts at 11.

Worked example 2: a shifted coordinate

Question. Find the path for x=1+2tx=1+2t, y=3t2y=3t^2, where 0≤t≤30\le t\le3.

Plan. Remove the horizontal offset before solving for time.

t=x−12remove offsety=34(x−1)2substitute0≤t≤3  ⟹  1≤x≤7shifted bounds\begin{aligned} t&=\frac{x-1}{2} && \text{remove offset}\\ y&=\frac34(x-1)^2 && \text{substitute}\\ 0\le t\le3&\implies1\le x\le7 && \text{shifted bounds} \end{aligned}

Why this step? Substituting t=x/2t=x/2 would forget the initial position.

Answer. y=34(x−1)2y=\frac34(x-1)^2, for 1≤x≤71\le x\le7.

Check. Its endpoints are (1,0)(1,0) and (7,27)(7,27), agreeing with t=0t=0 and 33.

There is no separate elimination formula in the booklet: reconstruct the method from the coordinate equations.

Watch out

Wrong: give the whole algebraic curve when time is restricted. Right: identify the travelled arc or state its coordinate bounds.

Use the same method with a new offset and scale.

Deriving a projectile trajectory

Horizontal motion supplies the time; vertical motion supplies the height.

Model the projectile as a particle with no air resistance, over distances small enough for gravity to have constant magnitude and direction. Take horizontal xx, upward yy, launch speed u>0u>0, and −π/2<θ<π/2-\pi/2<\theta<\pi/2, so horizontal velocity is positive.

Projectile coordinates and trajectory

With g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}},

x=ucos⁡θ t,y=usin⁡θ t−12gt2x=u\cos\theta\,t,\qquad y=u\sin\theta\,t-\frac12gt^2

Eliminating t=x/(ucos⁡θ)t=x/(u\cos\theta) gives

y=xtan⁡θ−gx22u2cos⁡2θy=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}

The coordinate equations follow from the booklet's constant-acceleration formulae. The trajectory is not printed separately.

Not in the formula booklet: learn it.

Derive the trajectory when asked to “show that”: quoting it skips the elimination.

Worked example 3: a forty-five-degree launch

Question. A particle is projected from level ground at 14 m s−114\,\mathrm{m\,s^{-1}}, at 45∘45^\circ. Derive its path until it next hits the ground.

Plan. Resolve the initial velocity, eliminate time and find the next ground intercept.

x=72 t,y=72 t−4.9t2velocity componentst=x72horizontal coordinatey=x−x220substitutey=0  ⟹  x(1−x/20)=0ground intercepts\begin{aligned} x&=7\sqrt2\,t,\quad y=7\sqrt2\,t-4.9t^2 && \text{velocity components}\\ t&=\frac{x}{7\sqrt2} && \text{horizontal coordinate}\\ y&=x-\frac{x^2}{20} && \text{substitute}\\ y=0&\implies x(1-x/20)=0 && \text{ground intercepts} \end{aligned}

Why this step? x=0x=0 is launch; x=20x=20 is the next contact.

Answer. y=x−x2/20y=x-x^2/20, for 0≤x≤200\le x\le20, with distances in metres.

Check. At the midpoint x=10x=10, height is 5 m5\,\mathrm{m}, the parabola's maximum.

Change only the angle to 30∘30^\circ, keeping total speed fixed.

Worked example 4: a lower angle

Question. Derive the path for u=14 m s−1u=14\,\mathrm{m\,s^{-1}}, θ=30∘\theta=30^\circ, until the next level-ground contact.

Plan. Resolve the new velocity and repeat the elimination.

x=73 t,y=7t−4.9t2new componentst=x73eliminate timey=x3−x230substitutey=0, x>0  ⟹  x=103next contact\begin{aligned} x&=7\sqrt3\,t,\quad y=7t-4.9t^2 && \text{new components}\\ t&=\frac{x}{7\sqrt3} && \text{eliminate time}\\ y&=\frac{x}{\sqrt3}-\frac{x^2}{30} && \text{substitute}\\ y=0,\ x>0&\implies x=10\sqrt3 && \text{next contact} \end{aligned}

Why this step? Changing angle changes both components; it does not keep horizontal speed fixed.

Answer. y=x/3−x2/30y=x/\sqrt3-x^2/30, for 0≤x≤1030\le x\le10\sqrt3.

Check. The range is about 17.3 m17.3\,\mathrm{m}, below the previous 20 m20\,\mathrm{m}.

The equations continue below ground algebraically. This free-flight model ends at the first subsequent ground contact.

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