MathsFurther Mechanics 2 › Further kinematics: acceleration as a function of x, t or v

Further kinematics: acceleration as a function of x, t or v

Two ways of writing the acceleration, and the whole skill is picking the one that separates. Get the pairing right and the integration is routine.

Builds on Newton's laws with a variable force and Solving differential equations.

IN THIS TOPIC

  • Choose the form of the acceleration that separates for a given problem.
  • Solve the resulting equation and apply the conditions correctly.
  • Interpret limiting behaviour, including terminal speed and total distance.

COMMON MISCONCEPTION

Acceleration is dv/dt, so any kinematics problem can be solved by integrating with respect to time.

Pick the pairing that separates

Acceleration has two equal forms, dv/dt and v dv/dx, and the choice between them is settled by the variable on the right-hand side together with the variable the question asks about:

dvdt=f(t) or f(v),vdvdx=f(v) or f(x)\frac{dv}{dt} = \text{f}(t) \text{ or f}(v), \qquad v\frac{dv}{dx} = \text{f}(v) \text{ or f}(x)NOT IN THE BOOKLET — LEARN IT

Neither form of the acceleration is in the booklet, so learn both and learn when each one separates.

Choose dv/dt when the acceleration depends on x and you are left with three variables in one equation and nowhere to go. Ask first what the acceleration depends on, then what the answer is wanted in terms of.

Once separated, integrate both sides and apply the conditions immediately, before any rearranging. If displacement is wanted from a velocity that depends on time, a second integration follows.

Resistance proportional to speed: v = 10 exp(−0.5t), and the area under the curve is the distance travelled17.3 m104 sdv/dt = −0.5vv = 1.35 m/s at t = 4
FIG. 1Speed against time under a resistance proportional to speed, with the distance travelled as the area beneath.

WORKED EXAMPLE

Resistance proportional to speed

A particle moving at 10 m/s decelerates at 0.5v m/s². Find its speed after 4 s and the distance it covers in that time.

The acceleration depends on v, and time is asked for, so use dv/dt = −0.5v.

Separating: ∫dv/v = −0.5∫dt gives ln v = −0.5t + c, and v = 10 at t = 0 gives v = 10e−0.5t.

At t = 4: v = 10e−2 = 1.35 m/s.

Distance = ∫v dt = 20(1 − e−2) = 17.3 m.

The same problem against distance

Change the question from 'after 4 seconds' to 'how far before it stops' and the identical physics needs the other pairing. Write the acceleration as v dv/dx, cancel the v, and a decaying exponential turns into a straight line. Two lines of work replace an integration by parts.

Read off the limiting behaviour while you are there. Under this resistance the speed approaches zero without reaching it in finite time, yet the total distance is finite at 20 m. Put a constant driving force against a resistance that grows with speed and the limit becomes a terminal speed instead, found by setting the acceleration to zero.

The same deceleration written against distance: v falls linearly and reaches zero at exactly 20 m20 m10v dv/dx = −0.5vso dv/dx = −0.5v = 10 − 0.5x
FIG. 2The same deceleration written against distance: a straight line reaching zero at exactly 20 m.

GUIDED PRACTICE

How far before it stops

For the same particle, find the total distance it travels before coming to rest, and comment.

Show the working

The acceleration depends on v and distance is wanted, so use v dv/dx = −0.5v.

Cancelling v, which is valid while the particle is moving: dv/dx = −0.5, so v = 10 − 0.5x.

v = 0 gives x = 20 m.

The particle never actually stops in finite time, since v decays exponentially, yet the distance it covers is finite. Both statements are true and neither contradicts the other.

ASSESSMENT FOCUS

  • State which form of the acceleration you are using and why, before separating anything.
  • Apply the initial conditions as soon as you have integrated, not after rearranging.
  • Watch for cancelling v. It is valid while the particle moves, and worth saying so.
  • For terminal speed, set the acceleration to zero instead of taking a limit.
  • Check the sign of a deceleration. Writing dv/dt = 0.5v instead of −0.5v turns a stopping particle into a runaway.

CHECK YOURSELF

A particle has acceleration 4t m/s² and speed 3 m/s at t = 0. Find its speed at t = 2.

Show a hint

The acceleration depends on t, so integrate with respect to t.

Show the answer

dv/dt = 4t, so v = 2t² + c, and v = 3 at t = 0 gives c = 3. At t = 2: v = 8 + 3 = 11 m/s.

Match the form of the acceleration to what it depends on: dv/dt for t or v, v dv/dx for x or v.

Separate, integrate, then apply the conditions at once.

A terminal speed is found by setting the acceleration to zero, not by taking a limit.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

6 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the further kinematics: acceleration as a function of x, t or v questions page.

CHECK YOUR PROGRESS

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  • Choose the form of the acceleration that separates for a given problem.
  • Solve the resulting equation and apply the conditions correctly.
  • Interpret limiting behaviour, including terminal speed and total distance.

Open the full revision checklist to see every objective in the course in one place.