4.2 - Constant acceleration

4.2 - Constant acceleration

At the highest point of a vertical throw, velocity is zero but gravitational acceleration is still acting. You will use the equations for constant acceleration, keep direction signs consistent, and connect stages or two particles with the correct times. These methods calculate motion without first needing to find the forces that caused it.

Choosing a motion equation

A particle starts at 6 m s−16\,\mathrm{m\,s^{-1}} and gains 2 m s−12\,\mathrm{m\,s^{-1}} each second. After 2 s2\,\mathrm{s}, its velocity is 10 m s−110\,\mathrm{m\,s^{-1}}. The straight velocity–time line has average height (6+10)/2=8(6+10)/2=8, so its area, and therefore displacement, is 16 m16\,\mathrm{m}.

This reasoning depends on constant acceleration: equal changes of velocity in equal times. It does not hold across an interval in which acceleration varies.

Constant-acceleration equations

For motion along a straight line with constant acceleration aa,

v=u+at,s=12(u+v)t,v=u+at,\qquad s=\tfrac12(u+v)t, s=ut+12at2,v2=u2+2as.s=ut+\tfrac12at^2,\qquad v^2=u^2+2as.

Here uu and vv are initial and final velocities, tt is elapsed time and ss is displacement during that interval, not final position unless the initial position is zero. Use one positive direction for s,u,v,as,u,v,a.

In the formula booklet.

The first equation comes from the definition of constant acceleration. The second is the trapezium area under its straight velocity graph. Substituting v=u+atv=u+at into that area gives s=ut+12at2s=ut+\tfrac12at^2; eliminating tt instead gives v2=u2+2asv^2=u^2+2as.

Choose an equation that includes what you need and excludes an unnecessary unknown.

Quantity not neededUseful equation
ssv=u+atv=u+at
aas=12(u+v)ts=\tfrac12(u+v)t
vvs=ut+12at2s=ut+\tfrac12at^2
ttv2=u2+2asv^2=u^2+2as

Write known signed values before substituting.

Worked example 1: positive acceleration

Question. A particle has initial velocity 6 m s−16\,\mathrm{m\,s^{-1}} and constant acceleration 2 m s−22\,\mathrm{m\,s^{-2}}. Find its velocity and displacement after 2 s2\,\mathrm{s}.

Plan. Take the initial motion as positive; use time in the velocity and displacement equations.

v=6+2(2)=10 m s−1s=6(2)+12(2)(22)=16 m\begin{aligned} v&=6+2(2)=10\,\mathrm{m\,s^{-1}}\\ s&=6(2)+\tfrac12(2)(2^2)=16\,\mathrm{m} \end{aligned}

Answer. Velocity 10 m s−110\,\mathrm{m\,s^{-1}}, displacement 16 m16\,\mathrm{m}.

Check. The velocity graph's area is 12(6+10)2=16 m\tfrac12(6+10)2=16\,\mathrm{m}, independently recovering the displacement.

Reverse only the acceleration. A negative aa is a signed change of velocity; it is not an instruction to make every other quantity negative.

Worked example 2: negative acceleration

Question. A particle has initial velocity 6 m s−16\,\mathrm{m\,s^{-1}} and constant acceleration −2 m s−2-2\,\mathrm{m\,s^{-2}}. Find its velocity and displacement after 2 s2\,\mathrm{s}.

Plan. Keep the initial direction positive and substitute a=−2a=-2.

v=6−2(2)=2 m s−1s=6(2)+12(−2)(22)=8 m\begin{aligned} v&=6-2(2)=2\,\mathrm{m\,s^{-1}}\\ s&=6(2)+\tfrac12(-2)(2^2)=8\,\mathrm{m} \end{aligned}

Answer. Velocity 2 m s−12\,\mathrm{m\,s^{-1}}, displacement 8 m8\,\mathrm{m}.

Check. Average velocity is (6+2)/2=4 m s−1(6+2)/2=4\,\mathrm{m\,s^{-1}}, giving displacement 4×2=8 m4\times2=8\,\mathrm{m}. Velocity stays positive over the interval, so distance is also 8 m8\,\mathrm{m}.

When using v2=u2+2asv^2=u^2+2as, the square gives the speed magnitude; decide the velocity sign from the stage of motion. For example, “first comes to rest” means v=0v=0, whereas reaching a height on the way down requires a negative velocity if up is positive.

Watch out

These equations require one constant acceleration over the chosen interval. If acceleration changes, divide the motion into stages and give each stage its own initial velocity and elapsed time.

Choose an equation that avoids an unknown time.

Vertical motion

For vertical motion with air resistance neglected, acceleration is constant and downwards. Taking up as positive gives a=−g=−10 m s−2a=-g=-10\,\mathrm{m\,s^{-2}}, during both the ascent and descent. At the top, v=0v=0 for an instant; aa has not become zero.

Use displacement relative to the launch point unless you explicitly define a ground-based position. A landing below the launch point has negative displacement when up is positive.

[DIAGRAM: asset_name: One acceleration through the turning point; asset_slug: mechanics-vertical-throw; caption: Velocity reverses at the top while acceleration remains downwards; description: A vertical position schematic shows a particle launched upwards from a 12 metre platform at 18 metres per second, rising 16.2 metres above launch and then falling to the ground. The top is labelled velocity zero and acceleration minus 10 metres per second squared. Up is positive, launch is position zero, and ground position minus 12 metres. Separate ascent and descent arrows identify 16.2 metres and 28.2 metres. A second panel plots velocity 18 minus 10t from launch to landing at 4.17 seconds, with the zero at 1.8 seconds and constant gradient minus 10.]
Velocity reverses at the top while acceleration remains downwards

Worked example 3: returning to the launch level

Question. A particle is projected vertically upwards from the ground at 18 m s−118\,\mathrm{m\,s^{-1}}. Neglect air resistance. Find its greatest height, time until it returns to the ground and total distance travelled by then.

Plan. Use v=0v=0 at the top, then s=0s=0 for the return to launch level.

02=182−20hhighest pointh=16.2 m0=18t−5t2=t(18−5t)return displacement zerot=3.6 sexclude the launch time 0\begin{aligned} 0^2&=18^2-20h &&\text{highest point}\\ h&=16.2\,\mathrm{m}\\ 0&=18t-5t^2=t(18-5t) &&\text{return displacement zero}\\ t&=3.6\,\mathrm{s} &&\text{exclude the launch time }0 \end{aligned}

Answer. Greatest height 16.2 m16.2\,\mathrm{m}; return time 3.6 s3.6\,\mathrm{s}; distance 2(16.2)=32.4 m2(16.2)=32.4\,\mathrm{m}.

Check. Time to the top from 0=18−10t0=18-10t is 1.8 s1.8\,\mathrm{s}, half the return time. The downward return velocity is 18−10(3.6)=−18 m s−118-10(3.6)=-18\,\mathrm{m\,s^{-1}}.

Raise only the launch point to a platform. The upward part is unchanged, but the descent is longer and the landing time is no longer twice the ascent time.

Worked example 4: landing below the launch point

Question. A particle is projected vertically upwards at 18 m s−118\,\mathrm{m\,s^{-1}} from a platform 12 m12\,\mathrm{m} above the ground. Neglect air resistance. Find its landing time and total distance travelled.

Plan. Take launch position as zero and up as positive; landing has displacement −12 m-12\,\mathrm{m}.

−12=18t−5t25t2−18t−12=0t=18+56410=4.17486… s\begin{aligned} -12&=18t-5t^2\\ 5t^2-18t-12&=0\\ t&=\frac{18+\sqrt{564}}{10}=4.17486\ldots\,\mathrm{s} \end{aligned}

The other root is negative, so it is outside the motion after projection. The ascent is still 16.2 m16.2\,\mathrm{m}. The descent to the ground is 16.2+12=28.2 m16.2+12=28.2\,\mathrm{m}.

Answer. Landing time 4.17 s4.17\,\mathrm{s}; total distance 16.2+28.2=44.4 m16.2+28.2=44.4\,\mathrm{m}.

Check. The landing velocity from v=18−10tv=18-10t is approximately −23.75 m s−1-23.75\,\mathrm{m\,s^{-1}}. Its square is 564564, agreeing with 182+2(−10)(−12)=56418^2+2(-10)(-12)=564.

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