Newton's Laws with Calculus

Force, mass, and acceleration relationships using derivatives and integrals

Newton's Laws with Calculus

Newton's Second Law

Vector form: F⃗net=ma⃗=mdv⃗dt=md2r⃗dt2\vec{F}_{net} = m\vec{a} = m\frac{d\vec{v}}{dt} = m\frac{d^2\vec{r}}{dt^2}

Component form: Fx=mdvxdt,Fy=mdvydt,Fz=mdvzdtF_x = m\frac{dv_x}{dt}, \quad F_y = m\frac{dv_y}{dt}, \quad F_z = m\frac{dv_z}{dt}

Impulse-Momentum Theorem

From Newton's second law: F⃗=dp⃗dt\vec{F} = \frac{d\vec{p}}{dt}

where momentum p⃗=mv⃗\vec{p} = m\vec{v}.

Integrating both sides: ∫t1t2F⃗ dt=∫t1t2dp⃗dt dt=p⃗2−p⃗1=Δp⃗\int_{t_1}^{t_2} \vec{F} \, dt = \int_{t_1}^{t_2} \frac{d\vec{p}}{dt} \, dt = \vec{p}_2 - \vec{p}_1 = \Delta \vec{p}

Impulse: J⃗=∫t1t2F⃗ dt=Δp⃗\vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt = \Delta \vec{p}

Variable Force Example

If F(t)=F0cos⁡(ωt)F(t) = F_0\cos(\omega t) for time 00 to T=π/(2ω)T = \pi/(2\omega):

J=∫0TF0cos⁡(ωt) dt=F0ωsin⁡(ωt)∣0T=F0ωJ = \int_0^T F_0\cos(\omega t) \, dt = \frac{F_0}{\omega}\sin(\omega t)\Big|_0^T = \frac{F_0}{\omega}

Variable Mass Systems

For systems where mass changes with time (rockets):

F⃗ext=dp⃗dt=d(mv⃗)dt=mdv⃗dt+v⃗dmdt\vec{F}_{ext} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt}

Rocket equation: mdvdt=−vreldmdt+Fextm\frac{dv}{dt} = -v_{rel}\frac{dm}{dt} + F_{ext}

where vrelv_{rel} is exhaust velocity relative to rocket.

For no external forces and integrating: Δv=vrelln⁡m0mf\Delta v = v_{rel}\ln\frac{m_0}{m_f}

(Tsiolkovsky rocket equation)

Drag Forces

Linear drag: Fd=−bvF_d = -bv

mdvdt=−bvm\frac{dv}{dt} = -bv

Solution: v(t)=v0e−bt/mv(t) = v_0e^{-bt/m}

Quadratic drag: Fd=−cv2F_d = -cv^2

mdvdt=−cv2m\frac{dv}{dt} = -cv^2

Separating variables: ∫v0vdv′v′2=−cm∫0tdt′\int_{v_0}^v \frac{dv'}{v'^2} = -\frac{c}{m}\int_0^t dt'

v(t)=v01+cv0tmv(t) = \frac{v_0}{1 + \frac{cv_0t}{m}}

Falling with Air Resistance

Equation of motion: mdvdt=mg−bvm\frac{dv}{dt} = mg - bv

Terminal velocity: vt=mgbv_t = \frac{mg}{b}

Rewrite as: dvdt=g−bmv=bm(vt−v)\frac{dv}{dt} = g - \frac{b}{m}v = \frac{b}{m}(v_t - v)

Solution: v(t)=vt(1−e−bt/m)v(t) = v_t(1 - e^{-bt/m})

Forces in Polar Coordinates

Newton's second law in polar coordinates:

Radial direction: Fr=m(r¨−rθ˙2)F_r = m(\ddot{r} - r\dot{\theta}^2)

Tangential direction: Fθ=m(rθ¨+2r˙θ˙)F_{\theta} = m(r\ddot{\theta} + 2\dot{r}\dot{\theta})

Example: Central Force

For motion under central force (depends only on rr):

Fθ=0F_{\theta} = 0, so: rθ¨+2r˙θ˙=0r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0

ddt(r2θ˙)=0\frac{d}{dt}(r^2\dot{\theta}) = 0

This means r2θ˙=L/mr^2\dot{\theta} = L/m is constant (angular momentum conservation).

📚 Practice Problems

1Problem 1medium

❓ Question:

A 2.0 kg block experiences a time-dependent force F(t) = (8 - 2t) N for 0 ≤ t ≤ 4 s. The block starts from rest. Find: (a) the impulse delivered to the block, (b) the final velocity, and (c) the work done by the force.

💡 Show Solution

Given:

  • m = 2.0 kg
  • F(t) = 8 - 2t N
  • v₀ = 0
  • Time interval: [0, 4] s

(a) Impulse:

J=∫04F(t) dt=∫04(8−2t) dtJ = \int_0^4 F(t) \, dt = \int_0^4 (8 - 2t) \, dt

J=[8t−t2]04=32−16J = \left[8t - t^2\right]_0^4 = 32 - 16

J=16 N\cdotps\boxed{J = 16 \text{ N·s}}

(b) Final velocity:

Using impulse-momentum theorem: J=Δp=m(vf−v0)J = \Delta p = m(v_f - v_0)

16=2.0(vf−0)16 = 2.0(v_f - 0)

vf=8.0 m/s\boxed{v_f = 8.0 \text{ m/s}}

(c) Work done:

First find position. Using F = ma: a(t)=Fm=8−2t2=4−ta(t) = \frac{F}{m} = \frac{8 - 2t}{2} = 4 - t

v(t)=∫0t(4−t′) dt′=4t−t22v(t) = \int_0^t (4 - t') \, dt' = 4t - \frac{t^2}{2}

x(t)=∫0t(4t′−t′22)dt′=2t2−t36x(t) = \int_0^t \left(4t' - \frac{t'^2}{2}\right) dt' = 2t^2 - \frac{t^3}{6}

Distance at t = 4: x(4)=2(16)−646=32−10.67=21.33 mx(4) = 2(16) - \frac{64}{6} = 32 - 10.67 = 21.33 \text{ m}

Work: W=∫04F(t)⋅v(t) dt=∫04(8−2t)(4t−t22)dtW = \int_0^4 F(t) \cdot v(t) \, dt = \int_0^4 (8-2t)\left(4t - \frac{t^2}{2}\right) dt

Or using work-energy theorem: W=ΔKE=12mvf2−0=12(2.0)(8.0)2W = \Delta KE = \frac{1}{2}mv_f^2 - 0 = \frac{1}{2}(2.0)(8.0)^2

W=64 J\boxed{W = 64 \text{ J}}

2Problem 2medium

❓ Question:

A 2.0 kg block experiences a time-dependent force F(t) = (8 - 2t) N for 0 ≤ t ≤ 4 s. The block starts from rest. Find: (a) the impulse delivered to the block, (b) the final velocity, and (c) the work done by the force.

💡 Show Solution

Given:

  • m = 2.0 kg
  • F(t) = 8 - 2t N
  • v₀ = 0
  • Time interval: [0, 4] s

(a) Impulse:

J=∫04F(t) dt=∫04(8−2t) dtJ = \int_0^4 F(t) \, dt = \int_0^4 (8 - 2t) \, dt

J=[8t−t2]04=32−16J = \left[8t - t^2\right]_0^4 = 32 - 16

J=16 N\cdotps\boxed{J = 16 \text{ N·s}}

(b) Final velocity:

Using impulse-momentum theorem: J=Δp=m(vf−v0)J = \Delta p = m(v_f - v_0)

16=2.0(vf−0)16 = 2.0(v_f - 0)

vf=8.0 m/s\boxed{v_f = 8.0 \text{ m/s}}

(c) Work done:

First find position. Using F = ma: a(t)=Fm=8−2t2=4−ta(t) = \frac{F}{m} = \frac{8 - 2t}{2} = 4 - t

v(t)=∫0t(4−t′) dt′=4t−t22v(t) = \int_0^t (4 - t') \, dt' = 4t - \frac{t^2}{2}

x(t)=∫0t(4t′−t′22)dt′=2t2−t36x(t) = \int_0^t \left(4t' - \frac{t'^2}{2}\right) dt' = 2t^2 - \frac{t^3}{6}

Distance at t = 4: x(4)=2(16)−646=32−10.67=21.33 mx(4) = 2(16) - \frac{64}{6} = 32 - 10.67 = 21.33 \text{ m}

Work: W=∫04F(t)⋅v(t) dt=∫04(8−2t)(4t−t22)dtW = \int_0^4 F(t) \cdot v(t) \, dt = \int_0^4 (8-2t)\left(4t - \frac{t^2}{2}\right) dt

Or using work-energy theorem: W=ΔKE=12mvf2−0=12(2.0)(8.0)2W = \Delta KE = \frac{1}{2}mv_f^2 - 0 = \frac{1}{2}(2.0)(8.0)^2

W=64 J\boxed{W = 64 \text{ J}}

3Problem 3medium

❓ Question:

A 2.0 kg block experiences a time-dependent force F(t) = (8 - 2t) N for 0 ≤ t ≤ 4 s. The block starts from rest. Find: (a) the impulse delivered to the block, (b) the final velocity, and (c) the work done by the force.

💡 Show Solution

Given:

  • m = 2.0 kg
  • F(t) = 8 - 2t N
  • v₀ = 0
  • Time interval: [0, 4] s

(a) Impulse:

J=∫04F(t) dt=∫04(8−2t) dtJ = \int_0^4 F(t) \, dt = \int_0^4 (8 - 2t) \, dt

J=[8t−t2]04=32−16J = \left[8t - t^2\right]_0^4 = 32 - 16

J=16 N\cdotps\boxed{J = 16 \text{ N·s}}

(b) Final velocity:

Using impulse-momentum theorem: J=Δp=m(vf−v0)J = \Delta p = m(v_f - v_0)

16=2.0(vf−0)16 = 2.0(v_f - 0)

vf=8.0 m/s\boxed{v_f = 8.0 \text{ m/s}}

(c) Work done:

First find position. Using F = ma: a(t)=Fm=8−2t2=4−ta(t) = \frac{F}{m} = \frac{8 - 2t}{2} = 4 - t

v(t)=∫0t(4−t′) dt′=4t−t22v(t) = \int_0^t (4 - t') \, dt' = 4t - \frac{t^2}{2}

x(t)=∫0t(4t′−t′22)dt′=2t2−t36x(t) = \int_0^t \left(4t' - \frac{t'^2}{2}\right) dt' = 2t^2 - \frac{t^3}{6}

Distance at t = 4: x(4)=2(16)−646=32−10.67=21.33 mx(4) = 2(16) - \frac{64}{6} = 32 - 10.67 = 21.33 \text{ m}

Work: W=∫04F(t)⋅v(t) dt=∫04(8−2t)(4t−t22)dtW = \int_0^4 F(t) \cdot v(t) \, dt = \int_0^4 (8-2t)\left(4t - \frac{t^2}{2}\right) dt

Or using work-energy theorem: W=ΔKE=12mvf2−0=12(2.0)(8.0)2W = \Delta KE = \frac{1}{2}mv_f^2 - 0 = \frac{1}{2}(2.0)(8.0)^2

W=64 J\boxed{W = 64 \text{ J}}

4Problem 4hard

❓ Question:

A rocket of initial mass m₀ = 1000 kg burns fuel at rate dm/dt = -10 kg/s with exhaust velocity v_e = 2000 m/s relative to the rocket. Neglecting gravity, find: (a) the thrust force, (b) the acceleration as a function of time, and (c) the velocity after 30 seconds.

💡 Show Solution

Given:

  • m₀ = 1000 kg
  • dm/dt = -10 kg/s
  • v_e = 2000 m/s
  • Ignore gravity

(a) Thrust force:

Rocket equation (thrust): Fthrust=−vedmdtF_{thrust} = -v_e \frac{dm}{dt}

Fthrust=−(2000)(−10)F_{thrust} = -(2000)(-10)

Fthrust=20,000 N=20 kN\boxed{F_{thrust} = 20,000 \text{ N} = 20 \text{ kN}}

(Constant as long as burn rate is constant)

(b) Acceleration as function of time:

Mass at time t: m(t)=m0+(dmdt)t=1000−10tm(t) = m_0 + \left(\frac{dm}{dt}\right)t = 1000 - 10t

Newton's second law: F=maF = ma 20000=(1000−10t)a20000 = (1000 - 10t)a

a(t)=200001000−10t=2000100−ta(t) = \frac{20000}{1000 - 10t} = \frac{2000}{100 - t}

a(t)=2000100−t m/s2\boxed{a(t) = \frac{2000}{100 - t} \text{ m/s}^2}

(c) Velocity after 30 s:

Using Tsiolkovsky rocket equation: Δv=veln⁡(m0mf)\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)

At t = 30 s: mf=1000−10(30)=700 kgm_f = 1000 - 10(30) = 700 \text{ kg}

Δv=2000ln⁡(1000700)=2000ln⁡(1.429)\Delta v = 2000 \ln\left(\frac{1000}{700}\right) = 2000 \ln(1.429)

Δv=2000(0.357)\Delta v = 2000(0.357)

v=714 m/s\boxed{v = 714 \text{ m/s}}

Alternative (integrate acceleration): v=∫030a(t) dt=∫0302000100−tdtv = \int_0^{30} a(t) \, dt = \int_0^{30} \frac{2000}{100-t} dt

v=−2000ln⁡(100−t)∣030=−2000[ln⁡(70)−ln⁡(100)]v = -2000\ln(100-t)\Big|_0^{30} = -2000[\ln(70) - \ln(100)]

v=−2000ln⁡(0.7)=2000(0.357)=714 m/sv = -2000\ln(0.7) = 2000(0.357) = 714 \text{ m/s} ✓

5Problem 5hard

❓ Question:

A rocket of initial mass m₀ = 1000 kg burns fuel at rate dm/dt = -10 kg/s with exhaust velocity v_e = 2000 m/s relative to the rocket. Neglecting gravity, find: (a) the thrust force, (b) the acceleration as a function of time, and (c) the velocity after 30 seconds.

💡 Show Solution

Given:

  • m₀ = 1000 kg
  • dm/dt = -10 kg/s
  • v_e = 2000 m/s
  • Ignore gravity

(a) Thrust force:

Rocket equation (thrust): Fthrust=−vedmdtF_{thrust} = -v_e \frac{dm}{dt}

Fthrust=−(2000)(−10)F_{thrust} = -(2000)(-10)

Fthrust=20,000 N=20 kN\boxed{F_{thrust} = 20,000 \text{ N} = 20 \text{ kN}}

(Constant as long as burn rate is constant)

(b) Acceleration as function of time:

Mass at time t: m(t)=m0+(dmdt)t=1000−10tm(t) = m_0 + \left(\frac{dm}{dt}\right)t = 1000 - 10t

Newton's second law: F=maF = ma 20000=(1000−10t)a20000 = (1000 - 10t)a

a(t)=200001000−10t=2000100−ta(t) = \frac{20000}{1000 - 10t} = \frac{2000}{100 - t}

a(t)=2000100−t m/s2\boxed{a(t) = \frac{2000}{100 - t} \text{ m/s}^2}

(c) Velocity after 30 s:

Using Tsiolkovsky rocket equation: Δv=veln⁡(m0mf)\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)

At t = 30 s: mf=1000−10(30)=700 kgm_f = 1000 - 10(30) = 700 \text{ kg}

Δv=2000ln⁡(1000700)=2000ln⁡(1.429)\Delta v = 2000 \ln\left(\frac{1000}{700}\right) = 2000 \ln(1.429)

Δv=2000(0.357)\Delta v = 2000(0.357)

v=714 m/s\boxed{v = 714 \text{ m/s}}

Alternative (integrate acceleration): v=∫030a(t) dt=∫0302000100−tdtv = \int_0^{30} a(t) \, dt = \int_0^{30} \frac{2000}{100-t} dt

v=−2000ln⁡(100−t)∣030=−2000[ln⁡(70)−ln⁡(100)]v = -2000\ln(100-t)\Big|_0^{30} = -2000[\ln(70) - \ln(100)]

v=−2000ln⁡(0.7)=2000(0.357)=714 m/sv = -2000\ln(0.7) = 2000(0.357) = 714 \text{ m/s} ✓

6Problem 6hard

❓ Question:

A rocket of initial mass m₀ = 1000 kg burns fuel at rate dm/dt = -10 kg/s with exhaust velocity v_e = 2000 m/s relative to the rocket. Neglecting gravity, find: (a) the thrust force, (b) the acceleration as a function of time, and (c) the velocity after 30 seconds.

💡 Show Solution

Given:

  • m₀ = 1000 kg
  • dm/dt = -10 kg/s
  • v_e = 2000 m/s
  • Ignore gravity

(a) Thrust force:

Rocket equation (thrust): Fthrust=−vedmdtF_{thrust} = -v_e \frac{dm}{dt}

Fthrust=−(2000)(−10)F_{thrust} = -(2000)(-10)

Fthrust=20,000 N=20 kN\boxed{F_{thrust} = 20,000 \text{ N} = 20 \text{ kN}}

(Constant as long as burn rate is constant)

(b) Acceleration as function of time:

Mass at time t: m(t)=m0+(dmdt)t=1000−10tm(t) = m_0 + \left(\frac{dm}{dt}\right)t = 1000 - 10t

Newton's second law: F=maF = ma 20000=(1000−10t)a20000 = (1000 - 10t)a

a(t)=200001000−10t=2000100−ta(t) = \frac{20000}{1000 - 10t} = \frac{2000}{100 - t}

a(t)=2000100−t m/s2\boxed{a(t) = \frac{2000}{100 - t} \text{ m/s}^2}

(c) Velocity after 30 s:

Using Tsiolkovsky rocket equation: Δv=veln⁡(m0mf)\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)

At t = 30 s: mf=1000−10(30)=700 kgm_f = 1000 - 10(30) = 700 \text{ kg}

Δv=2000ln⁡(1000700)=2000ln⁡(1.429)\Delta v = 2000 \ln\left(\frac{1000}{700}\right) = 2000 \ln(1.429)

Δv=2000(0.357)\Delta v = 2000(0.357)

v=714 m/s\boxed{v = 714 \text{ m/s}}

Alternative (integrate acceleration): v=∫030a(t) dt=∫0302000100−tdtv = \int_0^{30} a(t) \, dt = \int_0^{30} \frac{2000}{100-t} dt

v=−2000ln⁡(100−t)∣030=−2000[ln⁡(70)−ln⁡(100)]v = -2000\ln(100-t)\Big|_0^{30} = -2000[\ln(70) - \ln(100)]

v=−2000ln⁡(0.7)=2000(0.357)=714 m/sv = -2000\ln(0.7) = 2000(0.357) = 714 \text{ m/s} ✓

7Problem 7medium

❓ Question:

A 5.0 kg object falls through air with drag force F_d = bv², where b = 0.25 kg/m. Find: (a) the terminal velocity, (b) the differential equation for velocity, and (c) the time to reach 90% of terminal velocity (hint: separate variables and integrate).

💡 Show Solution

Given:

  • m = 5.0 kg
  • F_d = bv² where b = 0.25 kg/m
  • g = 9.8 m/s²

(a) Terminal velocity:

At terminal velocity, net force = 0: mg=bvt2mg = bv_t^2

vt=mgb=(5.0)(9.8)0.25v_t = \sqrt{\frac{mg}{b}} = \sqrt{\frac{(5.0)(9.8)}{0.25}}

vt=196v_t = \sqrt{196}

vt=14 m/s\boxed{v_t = 14 \text{ m/s}}

(b) Differential equation:

Newton's second law (taking down as positive): mdvdt=mg−bv2m\frac{dv}{dt} = mg - bv^2

dvdt=g−bmv2\boxed{\frac{dv}{dt} = g - \frac{b}{m}v^2}

Or: dvdt=g(1−v2vt2)\frac{dv}{dt} = g\left(1 - \frac{v^2}{v_t^2}\right)

(c) Time to reach 90% of v_t:

dvg(1−v2/vt2)=dt\frac{dv}{g(1 - v^2/v_t^2)} = dt

Using partial fractions or substitution u = v/v_t:

∫00.9vtdvg(1−v2/vt2)=∫0tdt\int_0^{0.9v_t} \frac{dv}{g(1 - v^2/v_t^2)} = \int_0^t dt

Result (hyperbolic tangent): vtgtanh⁡−1(vvt)=t\frac{v_t}{g}\tanh^{-1}\left(\frac{v}{v_t}\right) = t

At v = 0.9v_t: t=vtgtanh⁡−1(0.9)t = \frac{v_t}{g}\tanh^{-1}(0.9)

t=149.8tanh⁡−1(0.9)=(1.43)(1.472)t = \frac{14}{9.8}\tanh^{-1}(0.9) = (1.43)(1.472)

t=2.1 s\boxed{t = 2.1 \text{ s}}

Note: tanh⁡−1(0.9)=12ln⁡(1+0.91−0.9)=12ln⁡(19)=1.472\tanh^{-1}(0.9) = \frac{1}{2}\ln\left(\frac{1+0.9}{1-0.9}\right) = \frac{1}{2}\ln(19) = 1.472

8Problem 8medium

❓ Question:

A 5.0 kg object falls through air with drag force F_d = bv², where b = 0.25 kg/m. Find: (a) the terminal velocity, (b) the differential equation for velocity, and (c) the time to reach 90% of terminal velocity (hint: separate variables and integrate).

💡 Show Solution

Given:

  • m = 5.0 kg
  • F_d = bv² where b = 0.25 kg/m
  • g = 9.8 m/s²

(a) Terminal velocity:

At terminal velocity, net force = 0: mg=bvt2mg = bv_t^2

vt=mgb=(5.0)(9.8)0.25v_t = \sqrt{\frac{mg}{b}} = \sqrt{\frac{(5.0)(9.8)}{0.25}}

vt=196v_t = \sqrt{196}

vt=14 m/s\boxed{v_t = 14 \text{ m/s}}

(b) Differential equation:

Newton's second law (taking down as positive): mdvdt=mg−bv2m\frac{dv}{dt} = mg - bv^2

dvdt=g−bmv2\boxed{\frac{dv}{dt} = g - \frac{b}{m}v^2}

Or: dvdt=g(1−v2vt2)\frac{dv}{dt} = g\left(1 - \frac{v^2}{v_t^2}\right)

(c) Time to reach 90% of v_t:

dvg(1−v2/vt2)=dt\frac{dv}{g(1 - v^2/v_t^2)} = dt

Using partial fractions or substitution u = v/v_t:

∫00.9vtdvg(1−v2/vt2)=∫0tdt\int_0^{0.9v_t} \frac{dv}{g(1 - v^2/v_t^2)} = \int_0^t dt

Result (hyperbolic tangent): vtgtanh⁡−1(vvt)=t\frac{v_t}{g}\tanh^{-1}\left(\frac{v}{v_t}\right) = t

At v = 0.9v_t: t=vtgtanh⁡−1(0.9)t = \frac{v_t}{g}\tanh^{-1}(0.9)

t=149.8tanh⁡−1(0.9)=(1.43)(1.472)t = \frac{14}{9.8}\tanh^{-1}(0.9) = (1.43)(1.472)

t=2.1 s\boxed{t = 2.1 \text{ s}}

Note: tanh⁡−1(0.9)=12ln⁡(1+0.91−0.9)=12ln⁡(19)=1.472\tanh^{-1}(0.9) = \frac{1}{2}\ln\left(\frac{1+0.9}{1-0.9}\right) = \frac{1}{2}\ln(19) = 1.472

9Problem 9medium

❓ Question:

A 5.0 kg object falls through air with drag force F_d = bv², where b = 0.25 kg/m. Find: (a) the terminal velocity, (b) the differential equation for velocity, and (c) the time to reach 90% of terminal velocity (hint: separate variables and integrate).

💡 Show Solution

Given:

  • m = 5.0 kg
  • F_d = bv² where b = 0.25 kg/m
  • g = 9.8 m/s²

(a) Terminal velocity:

At terminal velocity, net force = 0: mg=bvt2mg = bv_t^2

vt=mgb=(5.0)(9.8)0.25v_t = \sqrt{\frac{mg}{b}} = \sqrt{\frac{(5.0)(9.8)}{0.25}}

vt=196v_t = \sqrt{196}

vt=14 m/s\boxed{v_t = 14 \text{ m/s}}

(b) Differential equation:

Newton's second law (taking down as positive): mdvdt=mg−bv2m\frac{dv}{dt} = mg - bv^2

dvdt=g−bmv2\boxed{\frac{dv}{dt} = g - \frac{b}{m}v^2}

Or: dvdt=g(1−v2vt2)\frac{dv}{dt} = g\left(1 - \frac{v^2}{v_t^2}\right)

(c) Time to reach 90% of v_t:

dvg(1−v2/vt2)=dt\frac{dv}{g(1 - v^2/v_t^2)} = dt

Using partial fractions or substitution u = v/v_t:

∫00.9vtdvg(1−v2/vt2)=∫0tdt\int_0^{0.9v_t} \frac{dv}{g(1 - v^2/v_t^2)} = \int_0^t dt

Result (hyperbolic tangent): vtgtanh⁡−1(vvt)=t\frac{v_t}{g}\tanh^{-1}\left(\frac{v}{v_t}\right) = t

At v = 0.9v_t: t=vtgtanh⁡−1(0.9)t = \frac{v_t}{g}\tanh^{-1}(0.9)

t=149.8tanh⁡−1(0.9)=(1.43)(1.472)t = \frac{14}{9.8}\tanh^{-1}(0.9) = (1.43)(1.472)

t=2.1 s\boxed{t = 2.1 \text{ s}}

Note: tanh⁡−1(0.9)=12ln⁡(1+0.91−0.9)=12ln⁡(19)=1.472\tanh^{-1}(0.9) = \frac{1}{2}\ln\left(\frac{1+0.9}{1-0.9}\right) = \frac{1}{2}\ln(19) = 1.472

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