Use acceleration forms and differential equations in rectilinear motion

Use the existing polynomial-motion foundation

Use displacement derivatives, rest and direction, distance and average speed as foundations. This lesson adds acceleration written in terms of position or velocity, differential-equation motion and branch/domain reasoning. Senior 6A/6B supply formulation, separation and valid solution intervals

Fix the direction and distinguish four quantities

Choose an oriented axis. Position x(t) and velocity v=dx/dt are signed; speed is |v|. Displacement over [t0, T] is x(T)-x(t0); distance is the integral of |v|. Acceleration a=dv/dt describes changing velocity. A negative velocity with positive acceleration is slowing while those signs persist

Choose an acceleration form by its independent variable

For time data, a=dv/dt=d²x/dt². On a differentiable branch v=V(x), the chain rule gives a=v dv/dx=d(v²/2)/dx. If the original acceleration is given as a function of v, solve dv/dt=a(v) first. Do not divide by zero velocity to force a position-based branch through a turn

Work a negative-velocity branch

For v(x)=-(2x+1), at x=1 the velocity is -3 and speed 3. The derivative dv/dx=-2, so acceleration is (-3)(-2)=6. The speed derivative is -6 because the velocity is negative. Acceleration has units m/s²; dv/dx alone has units inverse seconds

For v(x)=-(2x+1) at x=1, give the signed acceleration

What does this velocity–time graph imply?

Read areas and slopes with their signs

In the graph, the positive triangle has area 6 and the negative triangle has unsigned area 2. Displacement is 6-2=4m and distance is 6+2=8m. Over6s, average velocity is 2/3m/s and average speed is 4/3m/s. Acceleration is the segment slope away from corners:3/2,-3/2,-2, 2. At a sharp corner, a unique instantaneous acceleration is not defined by this piecewise-linear idealisation

A velocity–time graph joins (0,0), (2,3), (4,0), (5,-2) and (6,0) with straight segments, with time in seconds and velocity in m/s. Give the total distance in metres

Solve linear drag with either initial direction

For dv/dt=-kv with k>0 and v0≠0, separation gives v=v0e^(-kt). Then x=x0+(v0/k)(1-e^(-kt)). The sign of velocity stays fixed for finite forward time, so distance equals the magnitude of displacement. Speed tends to zero but never reaches zero at finite time; position tends to x0+v0/k. The separate initial value v0=0 gives stationary motion

Work the linear-drag initial conditions

For k=1/2, v0=-6, x0=2, v=-6e^(-t/2) and x=2-12(1-e^(-t/2)). At T=2ln3, velocity is -2, displacement is -8 and distance is 8. The limiting position is -10. Differentiating x recovers v, and differentiating v recovers -v/2.

For dv/dt=-v/2, v(0)=-6 and x(0)=2, give the limiting position as t tends to infinity

Use the velocity sign in quadratic drag

The sign-correct model dv/dt=-kv|v| opposes motion for either direction. On a nonzero-sign branch put s=sign(v0), w=|v| and w0=|v0|. Then w′=-kw², v=s w0/(1+kw0t), and x-x0=(s/k)ln(1+kw0t). The forward denominator is positive. The mathematical branch through 0 has t>-1/(kw0); v0=0 is a separate stationary solution

Compare two ideal drag laws

For k=1/4, v0=-4, x0=1, quadratic drag gives v=-4/(1+t), x=1-4ln(1+t). At t=3, v=-1 and distance is 4ln4. Speed tends to zero but the logarithmic distance grows without bound over infinite time. Linear drag has a finite limiting position instead. These are conclusions of the specified ideal laws, not universal claims about real resistance

For v(t)=-4/(1+t), give the velocity at t=3.

Integrate a position-dependent acceleration

For a=3x, x(0)=0 and v(0)=-2, use d(v²/2)/dx=3x. Integrating from the initial position gives v²=4+3x². On the branch moving toward negative x, choose v=-√(4+3x²). At x=-1, v²=7 and acceleration is -3. Squaring has not removed the need to choose the velocity sign

For a=3x, x(0)=0 and v(0)=-2, give v² when x=-1.

A terminal velocity can require a reversal

For dv/dt=k(U-v), the solution is v=U+(v0-U)e^(-kt). If U and v0 have opposite signs, a finite turn can occur. For U=2, k=1/2, v0=-4, x0=0, v=2-6e^(-t/2) and x=2t-12(1-e^(-t/2)). The unique zero is t*=2ln3. Velocity is negative before it and positive afterwards; the time equation is regular at the turn

Split distance at the turn

At t*=2ln3, position is 4ln3-8. At T=4ln3, position is 8ln3-32/3 and velocity 4/3. Total distance is -(4ln3-8)+[(8ln3-32/3)-(4ln3-8)]=16/3. Taking the absolute value of the final displacement alone would miss the return leg

For v(t)=2-6e^(-t/2), x(0)=0 and T=4ln3, give the total distance travelled from t=0 to T.

Rest need not mean a change of direction

For v=(t-2)² on [1, 3], velocity is positive on both sides of 2: the particle rests momentarily and continues in the same direction. For v=(t-2)³, signs are negative then positive, so it reverses even though acceleration also vanishes at 2. Multiplicity and a sign chart resolve the difference

For v=(t-2)³, what happens at t=2?

A restoring law can revisit the same position

For x″=-4x, x(0)=0 and v(0)=6, the solution x=3sin(2t) has v=6cos(2t). It satisfies both initial conditions and the equation. The first rest is at π/4, x=3; the next is at 3π/4, x=-3. The relation v²=36-4x² alone permits two velocity signs at most positions. Use a different branch on successive legs

Count only turns within the stated interval

On [0, 3π/4], the motion travels 0→3→-3, a distance 9, and ends at rest. Only the turn atπ/4 lies strictly inside the interval. The endpoint at 3π/4 is excluded from that interior count. Net displacement is -3, giving average velocity -4/π; average speed is 12/π. A future continuation would reverse at the endpoint, but that is outside this distance interval

Give the distance on [0, 3π/4] for x=3sin(2t)

A singular acceleration limits the solution

For dv/dt=-2/v, v(0)=-4 and x(0)=1, multiply by v only where v≠0. Integration gives v=-√(16-4t), with connected mathematical interval t<4. Position is 1-(64-(16-4t)^(3/2))/6. As t approaches4 from below, speed approaches zero but acceleration is undefined at that endpoint. A zero equilibrium is not a solution of the original equation

For dv/dt=-2/v with v(0)=-4, give the excluded upper time endpoint of the connected solution

Preserve initial conditions and model domains

Integrating acceleration with respect to time needs an initial velocity and then an initial position. Integrating d(v²/2)/dx needs the initial relation between position and velocity plus a branch sign. Check equilibria in the original equation, not only in a divided or multiplied form. At a turn, return to a regular time equation if the position representation ceases to be differentiable

Analyse and explain a velocity graph, ungraded

A velocity–time graph joins (0,0), (2,3), (4,0), (5,-2) and (6,0) with straight segments, with time in seconds and velocity in m/s. Use it to give displacement, distance, average velocity, average speed, direction intervals and acceleration on each open straight segment. Explain what its corners and endpoint zeros do and do not establish

Model: keep signed areas and derivative intervals separate

Signed areas are+6 and -2, giving displacement 4m and distance 8m. Divide by6s to obtain average velocity 2/3m/s and average speed 4/3m/s. Direction is positive on (0, 4) and negative on (4, 6), with one interior reversal at 4. Open-segment accelerations are 3/2,-3/2,-2, 2 m/s² on (0, 2),(2, 4),(4, 5),(5, 6). At2, 4, 5 the slopes jump, so the graph has no unique two-sided acceleration there. Endpoint zeros do not determine motion beyond the displayed interval. Position requires an additional initial position, though displacement does not

Derive and compare resistance models, ungraded

Derive signed velocity and position for both dv/dt=-kv and dv/dt=-kv|v|, k>0, with any nonzero initial velocity. Include v0=0 separately, state forward domains and explain finite versus infinite total distance. Identify what must be tested before applying either law to data

Model: solve on a sign branch and then integrate velocity

Linear drag gives v=v0e^(-kt), x=x0+(v0/k)(1-e^(-kt)); distance tends to |v0|/k. For quadratic drag put s=sign(v0), w0=|v0|. Separation gives v=s w0/(1+kw0t) and x=x0+(s/k)ln(1+kw0t), so distance diverges logarithmically as t→∞. Both have nonzero velocity at every finite forward time, preserve direction and admit stationary v=0 separately. Their coefficient units differ. Check resistance data, other forces, parameter stability and the interval of use; exact solutions verify the mathematics of the assumed laws rather than establish the laws empirically

Solve a reversal with terminal velocity, ungraded

For v′=(1/2)(2-v), v(0)=-4, x(0)=0, derive v and x. Find the turn, then displacement and distance to T=4ln3. State why crossing v=0 is permitted here and verify both initial conditions

Model: retain the signed exponential constant and split the path

Solving v′+(1/2)v=1 gives v=2-6e^(-t/2). Integrating and imposing x(0)=0 gives x=2t-12(1-e^(-t/2)). At zero, v=-4 and x=0. The turn t*=2ln3 has x*=4ln3-8<0. At T=4ln3, xT=8ln3-32/3 and vT=4/3. Distance is -x*+(xT-x*)=16/3, while displacement is xT. The right-hand side (1/2)(2-v) is defined and smooth at v=0, where acceleration is 1, so a regular reversal is possible

Explain velocity branches in oscillatory motion, ungraded

For x″=-4x, x(0)=0, v(0)=6, verify x=3sin(2t), derive v²=36-4x², and explain how the velocity sign changes across two successive turns. Determine distance and average speed over one complete period

Model: use the time solution through the turns

Differentiation gives v=6cos(2t), a=-12sin(2t)=-4x and the required initial data. Eliminating t yields v²=36-4x². The positive square root applies while moving right and the negative square root while moving left; a single-valued differentiable v(x) cannot describe the whole oscillation. The time solution remains smooth at x=±3. The period isπ, total distance is 4×3=12 and average speed is 12/π. Net displacement and average velocity over that full period are zero

Distinguish a limiting rest from an admissible endpoint, ungraded

For v′=-2/v, v(0)=-4, x(0)=1, derive the solution and maximal connected time interval. Find displacement and distance to t=3. Explain why finite limiting position and zero limiting speed do not allow the original equation to continue at t=4.

Model: check the original denominator after integration

From v dv/dt=-2 obtain v²=16-4t and select v=-√(16-4t). Then x=1-(64-(16-4t)^(3/2))/6. Its derivatives recover v and v′=-2/v wherever t<4; the physical forward interval is 0≤t<4. At t=3, v=-2, displacement is -28/3 and distance 28/3. As t→4−, position tends to -29/3, but v′ becomes unbounded and the original right-hand side is undefined. Adding stationary motion after 4 would change the model; it would not be a classical continuation of this equation

Assessment boundary

Practice grades bounded values, signs, area calculations, solution coefficients and specified domain classifications. Full derivations, motion explanations and empirical model validation remain ungraded