M1 3.1 - Constant acceleration

M1 3.1 - Constant acceleration

Choose and use the constant-acceleration equations with consistent signs. Apply them to vertical motion and to journeys split into separate stages, keeping distance distinct from signed displacement.

Choosing a constant-acceleration formula

A particle speeds up for a few seconds, then brakes to rest. Can one average speed describe the whole journey? Sometimes a weighted average can, but each stage has its own acceleration; treating them as one constant-acceleration interval usually fails.

Begin with one stage. Write uu for initial velocity, vv for final velocity, aa for constant acceleration, tt for elapsed time and ss for displacement during that interval. Choose one positive direction and use it for all three signed quantities u,v,au,v,a.

On a straight velocity-time graph, the gradient gives v=u+atv=u+at. The signed trapezium area gives s=12(u+v)ts=\tfrac12(u+v)t. Substituting for one variable produces the other useful equations.

Constant-acceleration equations

For a single interval of constant acceleration,

v=u+at,s=12(u+v)t,v=u+at,\qquad s=\frac12(u+v)t, s=ut+12at2,s=vt−12at2,v2=u2+2as.s=ut+\frac12at^2,\qquad s=vt-\frac12at^2,\qquad v^2=u^2+2as.

These are not in the formula booklet: learn them. Use compatible units, such as metres and seconds. If acceleration changes, start a new interval.

Choose the equation that includes the wanted quantity and your known quantities. For example, omit ss to find vv from u,a,tu,a,t; omit tt to find vv from u,a,su,a,s.

Worked example 1: time is known

Question. A particle has initial velocity 2 m s−12\,\mathrm{m\,s^{-1}} and constant acceleration 3 m s−23\,\mathrm{m\,s^{-2}}. Find its velocity after 4 s4\,\mathrm s.

Plan. Use v=u+atv=u+at, which does not need displacement.

v=2+3×4=14 m s−1.v=2+3\times4=14\,\mathrm{m\,s^{-1}}.

Answer. 14 m s−114\,\mathrm{m\,s^{-1}}.

Check. The average velocity is 88, so displacement is 32 m32\,\mathrm m. Substituting into v2=u2+2asv^2=u^2+2as gives 4+2×3×32=196=1424+2\times3\times32=196=14^2.

Change only the supplied interval measure: displacement replaces time.

Worked example 2: time is not known

Question. The same initial velocity and acceleration apply, but the given displacement is 32 m32\,\mathrm m. Find the final velocity.

Plan. Use v2=u2+2asv^2=u^2+2as, which does not need time; choose the sign consistent with the motion.

v2=22+2×3×32=196.v^2=2^2+2\times3\times32=196.

The particle begins moving positively and accelerates positively, so v>0v>0.

Answer. v=14 m s−1v=14\,\mathrm{m\,s^{-1}}.

Check. t=(14−2)/3=4 st=(14-2)/3=4\,\mathrm s, and 2×4+12×3×42=32 m2\times4+\tfrac12\times3\times4^2=32\,\mathrm m.

The square-root step needs a physical sign decision.

Watch out

v2=196v^2=196 alone gives two algebraic possibilities. Choose the velocity direction from the model. If the question asks for speed, report a nonnegative magnitude.

Choose an equation without finding an unnecessary time first.

Signed displacement velocity and acceleration

An acceleration opposite the velocity slows a particle initially, but it can later make the particle move back in the opposite direction.

A negative acceleration does not always mean slowing down. With right chosen positive, v<0v<0 and a<0a<0 mean the particle is moving left and gaining speed. Check the velocity as well as the acceleration.

Displacement through a reversal

The constant-acceleration equations give signed displacement even across a reversal. Distance must be split at the turning instant, where v=0v=0, and the positive lengths of the two legs added.

If u=12u=12 and a=−3a=-3, the velocity reaches zero after 44 seconds. An interval shorter than this has no reversal; a longer one does. This sign analysis comes before choosing whether distance equals the size of displacement.

Worked example 3: before the reversal

Question. A particle has initial velocity 12 m s−112\,\mathrm{m\,s^{-1}} and constant acceleration −3 m s−2-3\,\mathrm{m\,s^{-2}}. Find its velocity and displacement after 2 s2\,\mathrm s.

Plan. Use the same positive direction in both equations.

v=12−3×2=6 m s−1,v=12-3\times2=6\,\mathrm{m\,s^{-1}}, s=12×2−12×3×22=18 m.s=12\times2-\frac12\times3\times2^2=18\,\mathrm m.

Answer. Velocity 6 m s−16\,\mathrm{m\,s^{-1}}, displacement 18 m18\,\mathrm m.

Check. Average velocity (12+6)/2=9(12+6)/2=9; 9×2=189\times2=18. Velocity stays positive, so the distance is also 18 m18\,\mathrm m.

Change only the finishing time from 22 to 66 seconds. A reversal now lies inside the interval.

Worked example 4: after the reversal

Question. For the same initial velocity and acceleration, find velocity, displacement and distance travelled after 6 s6\,\mathrm s.

Plan. Find velocity and displacement normally, then split distance at the zero-velocity instant.

v=12−3×6=−6 m s−1,s=12×6−12×3×62=18 m.v=12-3\times6=-6\,\mathrm{m\,s^{-1}}, \qquad s=12\times6-\frac12\times3\times6^2=18\,\mathrm m.

The turning instant is t=12/3=4 st=12/3=4\,\mathrm s, when

sturn=12×4−12×3×42=24 m.s_{\mathrm{turn}}=12\times4-\frac12\times3\times4^2=24\,\mathrm m.

It then returns 24−18=6 m24-18=6\,\mathrm m.

Answer. Velocity −6 m s−1-6\,\mathrm{m\,s^{-1}}, displacement 18 m18\,\mathrm m, distance 24+6=30 m24+6=30\,\mathrm m.

Check. The velocity-time triangles give areas 2424 and −6-6: displacement 1818, distance 3030, confirming both results independently.

The same displacement can occur at two different times, with different velocity directions.

Vertical motion under gravity

A freely moving particle accelerates downwards both while rising and while falling.

Model the object as a particle and neglect air resistance. Over the distances considered, take gravitational acceleration as constant. With upwards positive, a=−ga=-g; with downwards positive, a=+ga=+g. In this lesson use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}. After substituting that numerical value, give final numerical answers to two or three significant figures, as Pearson's M1 papers instruct.

Key idea

At the highest point, vertical velocity is zero for an instant, but acceleration remains downward at gg. “Comes to rest” does not make gravity disappear.

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