Mechanics 7.3 - Constant acceleration formulae

Mechanics 7.3 - Constant acceleration formulae

Derive the constant-acceleration formulae from a velocity-time graph, then use them in straight-line and two-dimensional motion. You will choose equations, preserve vector components and distinguish displacement from position and speed.

Derive the straight-line formulae

How can a particle have the same displacement after 5 s5\,\mathrm s and 10 s10\,\mathrm s, yet travel further during the longer interval? Start with velocity 6 m s−16\,\mathrm{m\,s^{-1}} and constant acceleration −0.8 m s−2-0.8\,\mathrm{m\,s^{-2}}. A reversal will explain the puzzle.

You already know

A velocity-time gradient gives acceleration and signed area gives displacement. State a positive direction before assigning signs.

The five scalar quantities are displacement ss, initial velocity uu, final velocity vv, constant acceleration aa and elapsed time tt. They refer to the same interval. Constant acceleration means equal velocity changes in equal times, so the velocity-time graph is straight.

[DIAGRAM: asset_slug: p3_7_3_understand_and_use_formulae_for_constant_acceleration_including_2d__diagram_01; asset_name: Deriving constant-acceleration formulae; caption: Gradient gives velocity change and trapezium area gives displacement; description: A velocity-time line from u to v over time T has a rectangle uT and triangle one half times v minus u times T, with separately spaced derivation equations.]
Gradient gives velocity change and trapezium area gives displacement

Its gradient and signed area give

a=v−ut,s=12(u+v)t.a=\frac{v-u}{t},\qquad s=\tfrac12(u+v)t.

Thus v=u+atv=u+at. Substitution into the area gives s=ut+12at2s=ut+\tfrac12at^2; substituting u=v−atu=v-at instead gives s=vt−12at2s=vt-\tfrac12at^2. To eliminate time without dividing by aa, multiply the area equation by 2a2a:

2as=(u+v)at=(u+v)(v−u)=v2−u2.2as=(u+v)at=(u+v)(v-u)=v^2-u^2.

This also works when a=0a=0. Signed area keeps the derivation valid if the graph crosses the axis.

Straight-line constant acceleration

v=u+ats=12(u+v)ts=ut+12at2s=vt−12at2v2=u2+2as\begin{aligned} v&=u+at & s&=\tfrac12(u+v)t\\ s&=ut+\tfrac12at^2 & s&=vt-\tfrac12at^2\\ v^2&=u^2+2as \end{aligned}

Acceleration must be constant on the interval; all signed quantities use one positive direction. In the formula booklet.

The equations are consequences of one graph. Practise choosing one that avoids an unneeded unknown.

Worked example 1: eliminate time

Question. A particle has u=4 m s−1u=4\,\mathrm{m\,s^{-1}}, a=3 m s−2a=3\,\mathrm{m\,s^{-2}} and displacement s=8 ms=8\,\mathrm m. Find its final speed.

Plan. Use the equation without time.

Answer.

v2=42+2(3)(8)=64⇒∣v∣=8 m s−1.v^2=4^2+2(3)(8)=64\quad\Rightarrow\quad |v|=8\,\mathrm{m\,s^{-1}}.

Why take the positive square root? Speed is nonnegative. Here velocity is also positive because the acceleration increases an already positive velocity.

Check. t=(8−4)/3=4/3 st=(8-4)/3=4/3\,\mathrm s gives s=4(4/3)+12(3)(4/3)2=8 ms=4(4/3)+\tfrac12(3)(4/3)^2=8\,\mathrm m.

Change only the displacement from 88 to 14 m14\,\mathrm m.

Worked example 2: the same equation, further travel

Question. With the same u=4u=4 and a=3a=3, find final speed after displacement 14 m14\,\mathrm m.

Plan. Use v2=u2+2asv^2=u^2+2as again.

Answer.

v2=16+2(3)(14)=100⇒∣v∣=10 m s−1.v^2=16+2(3)(14)=100\quad\Rightarrow\quad |v|=10\,\mathrm{m\,s^{-1}}.

Why not find time first? It adds an unnecessary equation.

Check. t=(10−4)/3=2 st=(10-4)/3=2\,\mathrm s, and 4(2)+12(3)(22)=14 m4(2)+\tfrac12(3)(2^2)=14\,\mathrm m.

Use the same choice with a new displacement.

Use SUVAT in one dimension

List the known quantities and the requested one before choosing a formula. Signs describe directions; a negative acceleration does not force a negative displacement.

Worked example 3: before reversal

Question. A particle has initial velocity 6 m s−16\,\mathrm{m\,s^{-1}} and constant acceleration −0.8 m s−2-0.8\,\mathrm{m\,s^{-2}}. Find velocity and displacement after 5 s5\,\mathrm s.

Plan. Use the equations with known u,a,tu,a,t.

Answer.

v=6−0.8(5)=2 m s−1velocitys=6(5)−0.4(52)=20 mdisplacement\begin{aligned} v&=6-0.8(5)=2\,\mathrm{m\,s^{-1}} &&\text{velocity}\\ s&=6(5)-0.4(5^2)=20\,\mathrm m &&\text{displacement} \end{aligned}

Why retain the negative acceleration? It represents acceleration opposite to the chosen positive direction.

Check. Average velocity is (6+2)/2=4 m s−1(6+2)/2=4\,\mathrm{m\,s^{-1}}; multiplying by 5 s5\,\mathrm s gives 20 m20\,\mathrm m.

Change only the elapsed time to 10 s10\,\mathrm s, allowing the model to continue through the reversal.

Worked example 4: beyond reversal

Question. For the same particle, find velocity, displacement and total distance after 10 s10\,\mathrm s.

Plan. Find the turning time, then add the outward and return distances.

Answer.

v=6−0.8(10)=−2 m s−1s(10)=60−40=20 m0=6−0.8t⇒t=7.5 sturning times(7.5)=6(7.5)−0.4(7.5)2=22.5 mD=22.5+(22.5−20)=25 mtwo path lengths\begin{aligned} v&=6-0.8(10)=-2\,\mathrm{m\,s^{-1}}\\ s(10)&=60-40=20\,\mathrm m\\ 0&=6-0.8t\quad\Rightarrow\quad t=7.5\,\mathrm s &&\text{turning time}\\ s(7.5)&=6(7.5)-0.4(7.5)^2=22.5\,\mathrm m\\ D&=22.5+(22.5-20)=25\,\mathrm m &&\text{two path lengths} \end{aligned}

Why does the longer interval give the same displacement? The particle passes 20 m20\,\mathrm m on its way out and again on its way back.

Check. The signed trapezium area is 12(6−2)(10)=20 m\tfrac12(6-2)(10)=20\,\mathrm m, agreeing with displacement.

That is the opening puzzle resolved. In general, a constant nonzero acceleration does not stop acting just because velocity reaches zero; the particle then reverses if the model continues.

Watch out

Find any reversal before calling displacement a distance. Over an interval containing a reversal, add the magnitudes of the separate displacements.

Change only the elapsed time again.

Extend SUVAT to vectors

Two-dimensional motion uses the same scalar time in both perpendicular components. If u=4i−3j\mathbf u=4\mathbf i-3\mathbf j and a=2i+j\mathbf a=2\mathbf i+\mathbf j, then v=(4+2t)i+(−3+t)j\mathbf v=(4+2t)\mathbf i+(-3+t)\mathbf j. The two entries can equally be written as a column vector.

[DIAGRAM: asset_slug: p3_7_3_understand_and_use_formulae_for_constant_acceleration_including_2d__diagram_02; asset_name: Vector displacement and its components; caption: One vector calculation is two component calculations at the same time; description: A starting point A and later position P define displacement s with horizontal and vertical component legs, alongside separated component equations for vector constant acceleration.]
One vector calculation is two component calculations at the same time

Apply the scalar equations separately to the components and recombine them.

Vector constant acceleration

v=u+ats=ut+12at2=12(u+v)t=vt−12at2r=r0+s\begin{aligned} \mathbf v&=\mathbf u+\mathbf a t\\ \mathbf s&=\mathbf u t+\tfrac12\mathbf a t^2 =\tfrac12(\mathbf u+\mathbf v)t =\mathbf v t-\tfrac12\mathbf a t^2\\ \mathbf r&=\mathbf r_0+\mathbf s \end{aligned}

The vector a\mathbf a must be constant. These componentwise extensions and the position relation are not printed as vector formulae. Not in the formula booklet: learn it.

The scalar formula v2=u2+2asv^2=u^2+2as cannot be converted into a vector equation by making every symbol bold. Use components instead.

Worked example 5: vector velocity and displacement

Question. Given u=3i−2j m s−1\mathbf u=3\mathbf i-2\mathbf j\,\mathrm{m\,s^{-1}} and a=−0.4i+0.6j m s−2\mathbf a=-0.4\mathbf i+0.6\mathbf j\,\mathrm{m\,s^{-2}}, find velocity and displacement after 5 s5\,\mathrm s.

Plan. Multiply each component by the same scalar time.

Answer.

v=(3−0.4(5))i+(−2+0.6(5))j=i+j m s−1s=(3(5)−0.2(25))i+(−2(5)+0.3(25))j=10i−2.5j m\begin{aligned} \mathbf v&=(3-0.4(5))\mathbf i+(-2+0.6(5))\mathbf j =\mathbf i+\mathbf j\,\mathrm{m\,s^{-1}}\\ \mathbf s&=(3(5)-0.2(25))\mathbf i+(-2(5)+0.3(25))\mathbf j\\ &=10\mathbf i-2.5\mathbf j\,\mathrm m \end{aligned}

Why keep the components separate? They can have different signs and different accelerations.

Check. 12(u+v)5=10i−2.5j m\tfrac12(\mathbf u+\mathbf v)5=10\mathbf i-2.5\mathbf j\,\mathrm m.

Continue this lesson.

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