Why can't you use suvat equations with a variable force?

They assume constant acceleration.

Gravitational force at distance x from the Earth's centre?

\frac{mgR^2}{x^2} towards the centre.

At the surface \frac{GMm}{R^2}=mg.

Escape speed from the Earth's surface?

\sqrt{2gR}\approx1.12\times10^4 m s⁻¹.

What does separating variables mean?

Put all terms in v with dv on one side and all terms in the other variable on the other, then integrate.

A resistance kv acts. Does the particle reach rest in finite time?

No: v=ue^{-kt/m} is never zero. It does stop after a finite distance.

How do you find the constant of integration?

Substitute the initial conditions into the integrated equation.

Variable forces and motion in one dimensionEdexcel A-Level Further Maths: Flashcards

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Three forms of acceleration?

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Three forms of acceleration?: $\frac{dv}{dt}$ , $v\frac{dv}{dx}$ and $\frac{d^2x}{dt^2}$ .
Which form when $F$ depends on displacement $x$ ?: $mv\frac{dv}{dx}=F(x)$ .
Which form when $F$ depends on time $t$ ?: $m\frac{dv}{dt}=F(t)$ .
Why can't you use suvat equations with a variable force?: They assume constant acceleration.
Solve $\frac{dv}{dt}=-kv$ with $v=u$ at $t=0$ .: $v=ue^{-kt}$ .
Gravitational force at distance $x$ from the Earth's centre?: $\frac{mgR^2}{x^2}$ towards the centre.
Why is $GM=gR^2$ ?: At the surface $\frac{GMm}{R^2}=mg$ .
Speed at distance $x$ after projection upwards with speed $u$ ?: $v^2=u^2-2gR+\frac{2gR^2}{x}$ .
Escape speed from the Earth's surface?: $\sqrt{2gR}\approx1.12\times10^4$ m s⁻¹.
What does separating variables mean?: Put all terms in $v$ with $dv$ on one side and all terms in the other variable on the other, then integrate.
A resistance $kv$ acts. Does the particle reach rest in finite time?: No: $v=ue^{-kt/m}$ is never zero. It does stop after a finite distance.
How do you find the constant of integration?: Substitute the initial conditions into the integrated equation.

Exam questions on Variable forces and motion in one dimension

A particle of mass 2 kg moves in a straight line on a smooth horizontal surface. It starts from rest at the point $O$ and is acted on by a horizontal force of magnitude $6t$ newtons in the direction of motion, where $t$ is the time in seconds after the start.
Find the time at which the particle has speed 24 m s⁻¹.2 marks
A particle of mass 0.5 kg enters a viscous liquid with speed 8 m s⁻¹ and moves in a straight line. The only force acting on it in the direction of motion is a resistance of magnitude $2v$ newtons, where $v$ m s⁻¹ is its speed at time $t$ seconds after entering the liquid.
Find the distance travelled by the particle before it comes to rest.2 marks
A particle $P$ of mass 2 kg moves along the $x$ -axis on a smooth horizontal surface. When $P$ is at the point with coordinate $x$ metres, it is acted on by a force of magnitude $\frac{16}{(x+2)^2}$ newtons directed away from the origin $O$ . At $x=0$ the particle has speed 1 m s⁻¹ in the direction of increasing $x$ .
Show that $v^2=9-\frac{16}{x+2}$ , where $v$ m s⁻¹ is the speed of $P$ at the point with coordinate $x$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).