Motion with Variable Acceleration

Solving kinematics problems when acceleration depends on time, position, or velocity

Motion with Variable Acceleration

Acceleration as a Function of Time: a(t)

When acceleration varies with time:

v(t)=v0+∫t0ta(t′) dt′v(t) = v_0 + \int_{t_0}^t a(t') \, dt'

x(t)=x0+∫t0tv(t′) dt′x(t) = x_0 + \int_{t_0}^t v(t') \, dt'

Example: Linear Time Dependence

If a(t)=At+Ba(t) = At + B where AA and BB are constants:

v(t)=v0+∫0t(At′+B) dt′=v0+12At2+Btv(t) = v_0 + \int_0^t (At' + B) \, dt' = v_0 + \frac{1}{2}At^2 + Bt

x(t)=x0+∫0t(v0+12At′2+Bt′)dt′=x0+v0t+16At3+12Bt2x(t) = x_0 + \int_0^t \left(v_0 + \frac{1}{2}At'^2 + Bt'\right) dt' = x_0 + v_0t + \frac{1}{6}At^3 + \frac{1}{2}Bt^2

Acceleration as a Function of Position: a(x)

When acceleration depends on position, use the chain rule:

a=dvdt=dvdxdxdt=vdvdxa = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}

Therefore: v dv=a(x) dxv \, dv = a(x) \, dx

Integrating both sides: ∫v0vv′ dv′=∫x0xa(x′) dx′\int_{v_0}^v v' \, dv' = \int_{x_0}^x a(x') \, dx'

12(v2−v02)=∫x0xa(x′) dx′\frac{1}{2}(v^2 - v_0^2) = \int_{x_0}^x a(x') \, dx'

Example: Spring Force

For a spring with a=−kmxa = -\frac{k}{m}x:

v dv=−kmx dxv \, dv = -\frac{k}{m}x \, dx

∫v0vv′ dv′=−km∫x0xx′ dx′\int_{v_0}^v v' \, dv' = -\frac{k}{m}\int_{x_0}^x x' \, dx'

12(v2−v02)=−k2m(x2−x02)\frac{1}{2}(v^2 - v_0^2) = -\frac{k}{2m}(x^2 - x_0^2)

v2=v02−km(x2−x02)v^2 = v_0^2 - \frac{k}{m}(x^2 - x_0^2)

Acceleration as a Function of Velocity: a(v)

When acceleration depends on velocity:

a=dvdt=a(v)a = \frac{dv}{dt} = a(v)

Separate variables: dva(v)=dt\frac{dv}{a(v)} = dt

Integrate: ∫v0vdv′a(v′)=∫0tdt′=t\int_{v_0}^v \frac{dv'}{a(v')} = \int_0^t dt' = t

Example: Linear Drag Force

For drag force a=−bva = -bv (where b>0b > 0):

dvdt=−bv\frac{dv}{dt} = -bv

dvv=−b dt\frac{dv}{v} = -b \, dt

∫v0vdv′v′=−b∫0tdt′\int_{v_0}^v \frac{dv'}{v'} = -b\int_0^t dt'

ln⁡vv0=−bt\ln\frac{v}{v_0} = -bt

v(t)=v0e−btv(t) = v_0e^{-bt}

To find position, integrate velocity:

x(t)=∫0tv0e−bt′ dt′=v0b(1−e−bt)x(t) = \int_0^t v_0e^{-bt'} \, dt' = \frac{v_0}{b}(1 - e^{-bt})

Quadratic Drag Force

For drag proportional to v2v^2: a=−bv2a = -bv^2

dvdt=−bv2\frac{dv}{dt} = -bv^2

∫v0vdv′v′2=−b∫0tdt′\int_{v_0}^v \frac{dv'}{v'^2} = -b\int_0^t dt'

−1v+1v0=−bt-\frac{1}{v} + \frac{1}{v_0} = -bt

v(t)=v01+bv0tv(t) = \frac{v_0}{1 + bv_0t}

Falling with Air Resistance

Terminal velocity occurs when drag force equals gravitational force:

For linear drag: a=g−bva = g - bv

At terminal velocity: vt=gbv_t = \frac{g}{b}

General solution: v(t)=vt(1−e−bt)+v0e−btv(t) = v_t(1 - e^{-bt}) + v_0e^{-bt}

For quadratic drag: a=g−bv2a = g - bv^2

Terminal velocity: vt=gbv_t = \sqrt{\frac{g}{b}}

📚 Practice Problems

1Problem 1medium

❓ Question:

A particle starts from rest at t = 0 with acceleration a(t) = 6t m/s² (where t is in seconds). Find: (a) the velocity at t = 3 s, (b) the position at t = 3 s, and (c) the average velocity over the interval [0, 3] s.

💡 Show Solution

Given:

  • a(t) = 6t m/s²
  • v₀ = 0, x₀ = 0

(a) Velocity at t = 3 s:

v(t)=v0+∫0ta(t′) dt′=∫0t6t′ dt′v(t) = v_0 + \int_0^t a(t') \, dt' = \int_0^t 6t' \, dt'

v(t)=6⋅t22=3t2v(t) = 6 \cdot \frac{t^2}{2} = 3t^2

v(3)=3(3)2=27 m/sv(3) = 3(3)^2 = \boxed{27 \text{ m/s}}

(b) Position at t = 3 s:

x(t)=x0+∫0tv(t′) dt′=∫0t3t′2 dt′x(t) = x_0 + \int_0^t v(t') \, dt' = \int_0^t 3t'^2 \, dt'

x(t)=3⋅t33=t3x(t) = 3 \cdot \frac{t^3}{3} = t^3

x(3)=(3)3=27 mx(3) = (3)^3 = \boxed{27 \text{ m}}

(c) Average velocity:

vavg=x(3)−x(0)3−0=273v_{avg} = \frac{x(3) - x(0)}{3 - 0} = \frac{27}{3}

vavg=9 m/s\boxed{v_{avg} = 9 \text{ m/s}}

2Problem 2medium

❓ Question:

A particle starts from rest at t = 0 with acceleration a(t) = 6t m/s² (where t is in seconds). Find: (a) the velocity at t = 3 s, (b) the position at t = 3 s, and (c) the average velocity over the interval [0, 3] s.

💡 Show Solution

Given:

  • a(t) = 6t m/s²
  • v₀ = 0, x₀ = 0

(a) Velocity at t = 3 s:

v(t)=v0+∫0ta(t′) dt′=∫0t6t′ dt′v(t) = v_0 + \int_0^t a(t') \, dt' = \int_0^t 6t' \, dt'

v(t)=6⋅t22=3t2v(t) = 6 \cdot \frac{t^2}{2} = 3t^2

v(3)=3(3)2=27 m/sv(3) = 3(3)^2 = \boxed{27 \text{ m/s}}

(b) Position at t = 3 s:

x(t)=x0+∫0tv(t′) dt′=∫0t3t′2 dt′x(t) = x_0 + \int_0^t v(t') \, dt' = \int_0^t 3t'^2 \, dt'

x(t)=3⋅t33=t3x(t) = 3 \cdot \frac{t^3}{3} = t^3

x(3)=(3)3=27 mx(3) = (3)^3 = \boxed{27 \text{ m}}

(c) Average velocity:

vavg=x(3)−x(0)3−0=273v_{avg} = \frac{x(3) - x(0)}{3 - 0} = \frac{27}{3}

vavg=9 m/s\boxed{v_{avg} = 9 \text{ m/s}}

3Problem 3medium

❓ Question:

A particle starts from rest at t = 0 with acceleration a(t) = 6t m/s² (where t is in seconds). Find: (a) the velocity at t = 3 s, (b) the position at t = 3 s, and (c) the average velocity over the interval [0, 3] s.

💡 Show Solution

Given:

  • a(t) = 6t m/s²
  • v₀ = 0, x₀ = 0

(a) Velocity at t = 3 s:

v(t)=v0+∫0ta(t′) dt′=∫0t6t′ dt′v(t) = v_0 + \int_0^t a(t') \, dt' = \int_0^t 6t' \, dt'

v(t)=6⋅t22=3t2v(t) = 6 \cdot \frac{t^2}{2} = 3t^2

v(3)=3(3)2=27 m/sv(3) = 3(3)^2 = \boxed{27 \text{ m/s}}

(b) Position at t = 3 s:

x(t)=x0+∫0tv(t′) dt′=∫0t3t′2 dt′x(t) = x_0 + \int_0^t v(t') \, dt' = \int_0^t 3t'^2 \, dt'

x(t)=3⋅t33=t3x(t) = 3 \cdot \frac{t^3}{3} = t^3

x(3)=(3)3=27 mx(3) = (3)^3 = \boxed{27 \text{ m}}

(c) Average velocity:

vavg=x(3)−x(0)3−0=273v_{avg} = \frac{x(3) - x(0)}{3 - 0} = \frac{27}{3}

vavg=9 m/s\boxed{v_{avg} = 9 \text{ m/s}}

4Problem 4hard

❓ Question:

A rocket experiences acceleration a = (40 - 5t) m/s² until its fuel runs out. The rocket starts from rest at t = 0. Find: (a) when the fuel runs out (when a = 0), (b) the maximum velocity, and (c) the total distance traveled while fuel is burning.

💡 Show Solution

Given:

  • a(t) = 40 - 5t m/s²
  • v₀ = 0, x₀ = 0

(a) When fuel runs out:

Set a = 0: 40−5t=040 - 5t = 0

t=8 s\boxed{t = 8 \text{ s}}

(b) Maximum velocity:

v(t)=∫0t(40−5t′) dt′=40t−5t22v(t) = \int_0^t (40 - 5t') \, dt' = 40t - \frac{5t^2}{2}

At t = 8 s: v(8)=40(8)−5(64)2=320−160v(8) = 40(8) - \frac{5(64)}{2} = 320 - 160

vmax=160 m/s\boxed{v_{max} = 160 \text{ m/s}}

(c) Total distance:

x(t)=∫0t(40t′−5t′22)dt′x(t) = \int_0^t \left(40t' - \frac{5t'^2}{2}\right) dt'

x(t)=40⋅t22−5t36=20t2−5t36x(t) = 40 \cdot \frac{t^2}{2} - \frac{5t^3}{6} = 20t^2 - \frac{5t^3}{6}

At t = 8 s: x(8)=20(64)−5(512)6=1280−426.7x(8) = 20(64) - \frac{5(512)}{6} = 1280 - 426.7

x=853 m\boxed{x = 853 \text{ m}}

5Problem 5hard

❓ Question:

A rocket experiences acceleration a = (40 - 5t) m/s² until its fuel runs out. The rocket starts from rest at t = 0. Find: (a) when the fuel runs out (when a = 0), (b) the maximum velocity, and (c) the total distance traveled while fuel is burning.

💡 Show Solution

Given:

  • a(t) = 40 - 5t m/s²
  • v₀ = 0, x₀ = 0

(a) When fuel runs out:

Set a = 0: 40−5t=040 - 5t = 0

t=8 s\boxed{t = 8 \text{ s}}

(b) Maximum velocity:

v(t)=∫0t(40−5t′) dt′=40t−5t22v(t) = \int_0^t (40 - 5t') \, dt' = 40t - \frac{5t^2}{2}

At t = 8 s: v(8)=40(8)−5(64)2=320−160v(8) = 40(8) - \frac{5(64)}{2} = 320 - 160

vmax=160 m/s\boxed{v_{max} = 160 \text{ m/s}}

(c) Total distance:

x(t)=∫0t(40t′−5t′22)dt′x(t) = \int_0^t \left(40t' - \frac{5t'^2}{2}\right) dt'

x(t)=40⋅t22−5t36=20t2−5t36x(t) = 40 \cdot \frac{t^2}{2} - \frac{5t^3}{6} = 20t^2 - \frac{5t^3}{6}

At t = 8 s: x(8)=20(64)−5(512)6=1280−426.7x(8) = 20(64) - \frac{5(512)}{6} = 1280 - 426.7

x=853 m\boxed{x = 853 \text{ m}}

6Problem 6hard

❓ Question:

A rocket experiences acceleration a = (40 - 5t) m/s² until its fuel runs out. The rocket starts from rest at t = 0. Find: (a) when the fuel runs out (when a = 0), (b) the maximum velocity, and (c) the total distance traveled while fuel is burning.

💡 Show Solution

Given:

  • a(t) = 40 - 5t m/s²
  • v₀ = 0, x₀ = 0

(a) When fuel runs out:

Set a = 0: 40−5t=040 - 5t = 0

t=8 s\boxed{t = 8 \text{ s}}

(b) Maximum velocity:

v(t)=∫0t(40−5t′) dt′=40t−5t22v(t) = \int_0^t (40 - 5t') \, dt' = 40t - \frac{5t^2}{2}

At t = 8 s: v(8)=40(8)−5(64)2=320−160v(8) = 40(8) - \frac{5(64)}{2} = 320 - 160

vmax=160 m/s\boxed{v_{max} = 160 \text{ m/s}}

(c) Total distance:

x(t)=∫0t(40t′−5t′22)dt′x(t) = \int_0^t \left(40t' - \frac{5t'^2}{2}\right) dt'

x(t)=40⋅t22−5t36=20t2−5t36x(t) = 40 \cdot \frac{t^2}{2} - \frac{5t^3}{6} = 20t^2 - \frac{5t^3}{6}

At t = 8 s: x(8)=20(64)−5(512)6=1280−426.7x(8) = 20(64) - \frac{5(512)}{6} = 1280 - 426.7

x=853 m\boxed{x = 853 \text{ m}}

7Problem 7hard

❓ Question:

A particle moves with position-dependent acceleration a = -kx, where k = 4 s⁻². If v = 8 m/s when x = 0, find: (a) the velocity as a function of position, (b) the maximum displacement, and (c) identify the type of motion.

💡 Show Solution

Given:

  • a = -kx where k = 4 s⁻²
  • At x = 0: v = 8 m/s

(a) Velocity as function of position:

Using a=vdvdxa = v\frac{dv}{dx}:

vdvdx=−kxv\frac{dv}{dx} = -kx

v dv=−kx dxv \, dv = -kx \, dx

Integrating: v22=−kx22+C\frac{v^2}{2} = -\frac{kx^2}{2} + C

At x = 0, v = 8: C=642=32C = \frac{64}{2} = 32

v22=−4x22+32\frac{v^2}{2} = -\frac{4x^2}{2} + 32

v2=64−4x2v^2 = 64 - 4x^2

v=64−4x2=216−x2 m/s\boxed{v = \sqrt{64 - 4x^2} = 2\sqrt{16 - x^2} \text{ m/s}}

(b) Maximum displacement:

At maximum displacement, v = 0: 0=64−4xmax20 = 64 - 4x_{max}^2

xmax2=16x_{max}^2 = 16

xmax=4 m\boxed{x_{max} = 4 \text{ m}}

(c) Type of motion:

a=−kx=−(4)xa = -kx = -(4)x

This is Simple Harmonic Motion (SHM)!

ω2=k=4  ⟹  ω=2 rad/s\omega^2 = k = 4 \implies \omega = 2 \text{ rad/s}

The particle oscillates with amplitude A = 4 m and angular frequency ω = 2 rad/s.

General solution: x(t)=Asin⁡(ωt+ϕ)x(t) = A\sin(\omega t + \phi)

8Problem 8hard

❓ Question:

A particle moves with position-dependent acceleration a = -kx, where k = 4 s⁻². If v = 8 m/s when x = 0, find: (a) the velocity as a function of position, (b) the maximum displacement, and (c) identify the type of motion.

💡 Show Solution

Given:

  • a = -kx where k = 4 s⁻²
  • At x = 0: v = 8 m/s

(a) Velocity as function of position:

Using a=vdvdxa = v\frac{dv}{dx}:

vdvdx=−kxv\frac{dv}{dx} = -kx

v dv=−kx dxv \, dv = -kx \, dx

Integrating: v22=−kx22+C\frac{v^2}{2} = -\frac{kx^2}{2} + C

At x = 0, v = 8: C=642=32C = \frac{64}{2} = 32

v22=−4x22+32\frac{v^2}{2} = -\frac{4x^2}{2} + 32

v2=64−4x2v^2 = 64 - 4x^2

v=64−4x2=216−x2 m/s\boxed{v = \sqrt{64 - 4x^2} = 2\sqrt{16 - x^2} \text{ m/s}}

(b) Maximum displacement:

At maximum displacement, v = 0: 0=64−4xmax20 = 64 - 4x_{max}^2

xmax2=16x_{max}^2 = 16

xmax=4 m\boxed{x_{max} = 4 \text{ m}}

(c) Type of motion:

a=−kx=−(4)xa = -kx = -(4)x

This is Simple Harmonic Motion (SHM)!

ω2=k=4  ⟹  ω=2 rad/s\omega^2 = k = 4 \implies \omega = 2 \text{ rad/s}

The particle oscillates with amplitude A = 4 m and angular frequency ω = 2 rad/s.

General solution: x(t)=Asin⁡(ωt+ϕ)x(t) = A\sin(\omega t + \phi)

9Problem 9hard

❓ Question:

A particle moves with position-dependent acceleration a = -kx, where k = 4 s⁻². If v = 8 m/s when x = 0, find: (a) the velocity as a function of position, (b) the maximum displacement, and (c) identify the type of motion.

💡 Show Solution

Given:

  • a = -kx where k = 4 s⁻²
  • At x = 0: v = 8 m/s

(a) Velocity as function of position:

Using a=vdvdxa = v\frac{dv}{dx}:

vdvdx=−kxv\frac{dv}{dx} = -kx

v dv=−kx dxv \, dv = -kx \, dx

Integrating: v22=−kx22+C\frac{v^2}{2} = -\frac{kx^2}{2} + C

At x = 0, v = 8: C=642=32C = \frac{64}{2} = 32

v22=−4x22+32\frac{v^2}{2} = -\frac{4x^2}{2} + 32

v2=64−4x2v^2 = 64 - 4x^2

v=64−4x2=216−x2 m/s\boxed{v = \sqrt{64 - 4x^2} = 2\sqrt{16 - x^2} \text{ m/s}}

(b) Maximum displacement:

At maximum displacement, v = 0: 0=64−4xmax20 = 64 - 4x_{max}^2

xmax2=16x_{max}^2 = 16

xmax=4 m\boxed{x_{max} = 4 \text{ m}}

(c) Type of motion:

a=−kx=−(4)xa = -kx = -(4)x

This is Simple Harmonic Motion (SHM)!

ω2=k=4  ⟹  ω=2 rad/s\omega^2 = k = 4 \implies \omega = 2 \text{ rad/s}

The particle oscillates with amplitude A = 4 m and angular frequency ω = 2 rad/s.

General solution: x(t)=Asin⁡(ωt+ϕ)x(t) = A\sin(\omega t + \phi)

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