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Resisted motion and projectiles

Model motion with resistance proportional to velocity (R=mkvR = mkv) or v2v^2 (R=mkv2R = mkv^2); solve the resulting separable differential equations; find and interpret terminal velocity; model projectile motion incorporating air resistance.

Worked examples

Horizontal motion with resistance proportional to velocity

Straightforward

Problem

A particle of mass 33 kg moves horizontally. The only horizontal force is air resistance R=6vR = 6v N (opposing motion). Given v=15v = 15 m/s at t=0t = 0, find (a) vv as a function of tt, and (b) the distance travelled until the particle's speed halves.

Falling body with resistance, finding terminal velocity and motion

Moderate

Problem

A particle of mass 11 kg falls from rest. Air resistance is R=0.2vR = 0.2v N (taking downward as positive, g=10g = 10 m/s²). Find the terminal velocity and show that v=50(1−e−0.2t)v = 50(1-e^{-0.2t}).

Resistance proportional to v2v^2 — using the vv–xx form

Challenging

Problem

A particle of mass 22 kg falls from rest with resistance R=0.08v2R = 0.08v^2 N. Using g=10g = 10 m/s², find the speed after falling 2020 m.

Practise

Q1·Straightforward
A particle of mass 22 kg moves horizontally with a resistance force R=4vR = 4v N (opposing motion). Given initial velocity v0=10v_0 = 10 m/s, find the velocity (in m/s) at t=1t = 1 s. Give your answer to two decimal places.
Q2·Straightforward
A particle satisfies v˙=−3v\dot{v} = -3v with v(0)=12v(0) = 12 m/s. Find the velocity (in m/s) when t=0.5t = 0.5 s. Give your answer to two decimal places.
Q3·Straightforward
A particle of mass 22 kg falls from rest under gravity with air resistance R=2vR = 2v N. Using g=10g = 10 m/s², find the terminal velocity (in m/s).
Q4·Straightforward
A particle of mass 11 kg falls with air resistance R=0.1v2R = 0.1v^2 N (taking downward as positive, g=10g = 10 m/s²). Find the terminal velocity (in m/s).
Q5·Moderate
A particle of mass 11 kg falls from rest with equation of motion v˙=10−2v\dot{v} = 10 - 2v (taking downward as positive, g=10g = 10 m/s²). Find the velocity (in m/s) at t=1t = 1 s. Give your answer to two decimal places.
Q6·Moderate
A 55 kg particle moves horizontally with resistance R=5v2R = 5v^2 N and no driving force. At t=0t = 0, v=20v = 20 m/s. Find the time (in seconds) when v=10v = 10 m/s. Give your answer to three decimal places.
Q7·Moderate
A particle of mass 11 kg falls from rest with equation of motion v˙=10−0.5v\dot{v} = 10 - 0.5v (downward positive). Find the distance fallen (in metres) in the first 22 s. Give your answer to two decimal places.
Q8·Straightforward
A falling particle satisfies v˙=10−0.2v\dot{v} = 10 - 0.2v (with g=10g = 10 m/s²). Find the terminal velocity (in m/s).
Q9·Moderate
A particle of mass 11 kg moves horizontally with a resistance force R=0.1v2R = 0.1v^2 N. Its initial velocity is 2020 m/s. Find the distance (in metres) the particle travels until its velocity falls to 55 m/s. Give your answer to two decimal places.
Q10·Straightforward
A particle moves horizontally. Its velocity satisfies dvdt=−0.5v\dfrac{dv}{dt} = -0.5v with v(0)=10v(0) = 10 m/s. Find the velocity (in m/s) at t=4t = 4 s. Give your answer to two decimal places.
Q11·Challenging
A particle of mass 11 kg falls with v˙=10−0.4v2\dot{v} = 10 - 0.4v^2 (downward positive, g=10g = 10 m/s²). Find the time (in seconds) to reach half the terminal velocity. Give your answer to two decimal places.
Q12·Challenging
A particle of mass mm falls from rest under gravity with resistance R=mkvR = mkv (proportional to velocity, k>0k > 0). Prove that v=gk(1−e−kt)v = \dfrac{g}{k}\left(1 - e^{-kt}\right) and state the terminal velocity.

✎ Work this one through on paper — proofs are self-assessed.