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Position, Velocity, and Acceleration

Derivatives and integrals in kinematics

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Position, Velocity, and Acceleration

Fundamental Relationships

Position, velocity, and acceleration are related through calculus:

Velocity as derivative of position: v⃗(t)=dr⃗dt\vec{v}(t) = \frac{d\vec{r}}{dt}

Acceleration as derivative of velocity: a⃗(t)=dv⃗dt=d2r⃗dt2\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}

Position from velocity (integration): r⃗(t)=r⃗0+∫t0tv⃗(t′) dt′\vec{r}(t) = \vec{r}_0 + \int_{t_0}^{t} \vec{v}(t') \, dt'

Velocity from acceleration (integration): v⃗(t)=v⃗0+∫t0ta⃗(t′) dt′\vec{v}(t) = \vec{v}_0 + \int_{t_0}^{t} \vec{a}(t') \, dt'

One-Dimensional Motion

For motion along a straight line (x-axis):

Position function: x(t)x(t)

Velocity: vx(t)=dxdtv_x(t) = \frac{dx}{dt}

Acceleration: ax(t)=dvxdt=d2xdt2a_x(t) = \frac{dv_x}{dt} = \frac{d^2x}{dt^2}

Example: Polynomial Position Function

If x(t)=3t3−2t2+5t+1x(t) = 3t^3 - 2t^2 + 5t + 1 (in meters), then:

Velocity: vx=dxdt=9t2−4t+5 m/sv_x = \frac{dx}{dt} = 9t^2 - 4t + 5 \text{ m/s}

Acceleration: ax=dvxdt=18t−4 m/s2a_x = \frac{dv_x}{dt} = 18t - 4 \text{ m/s}^2

Two-Dimensional Motion

Position vector in 2D: r⃗(t)=x(t)i^+y(t)j^\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}

Velocity vector: v⃗(t)=dxdti^+dydtj^=vxi^+vyj^\vec{v}(t) = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} = v_x\hat{i} + v_y\hat{j}

Acceleration vector: a⃗(t)=dvxdti^+dvydtj^=axi^+ayj^\vec{a}(t) = \frac{dv_x}{dt}\hat{i} + \frac{dv_y}{dt}\hat{j} = a_x\hat{i} + a_y\hat{j}

Speed (magnitude of velocity): v=∣v⃗∣=vx2+vy2v = |\vec{v}| = \sqrt{v_x^2 + v_y^2}

Projectile Motion

For projectile motion with no air resistance:

Horizontal component:

  • ax=0a_x = 0
  • vx=v0x=v0cos⁡θv_x = v_{0x} = v_0\cos\theta (constant)
  • x(t)=x0+v0xtx(t) = x_0 + v_{0x}t

Vertical component:

  • ay=−ga_y = -g
  • vy(t)=v0y−gt=v0sin⁡θ−gtv_y(t) = v_{0y} - gt = v_0\sin\theta - gt
  • y(t)=y0+v0yt−12gt2y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2

Variable Acceleration

When acceleration is a function of time, a(t)a(t):

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

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

Example: Time-Dependent Acceleration

If a(t)=6t−4a(t) = 6t - 4 m/s², and v0=2v_0 = 2 m/s, x0=0x_0 = 0:

v(t)=2+∫0t(6t′−4) dt′=2+3t2−4tv(t) = 2 + \int_0^t (6t' - 4) \, dt' = 2 + 3t^2 - 4t

x(t)=∫0t(2+3t′2−4t′) dt′=2t+t3−2t2x(t) = \int_0^t (2 + 3t'^2 - 4t') \, dt' = 2t + t^3 - 2t^2

📚 Practice Problems

1Problem 1medium

❓ Question:

A particle moves along the x-axis with position given by x(t) = 2t³ - 9t² + 12t + 5, where x is in meters and t is in seconds. Find: (a) the velocity and acceleration as functions of time, (b) the times when the particle is at rest, and (c) the position when the acceleration is zero.

💡 Show Solution

Given: x(t)=2t3−9t2+12t+5x(t) = 2t^3 - 9t^2 + 12t + 5

(a) Velocity and acceleration:

Velocity: v(t)=dxdt=6t2−18t+12v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12

v(t)=6t2−18t+12 m/s\boxed{v(t) = 6t^2 - 18t + 12 \text{ m/s}}

Acceleration: a(t)=dvdt=12t−18a(t) = \frac{dv}{dt} = 12t - 18

a(t)=12t−18 m/s2\boxed{a(t) = 12t - 18 \text{ m/s}^2}

(b) Times when particle is at rest:

Set v(t) = 0: 6t2−18t+12=06t^2 - 18t + 12 = 0 t2−3t+2=0t^2 - 3t + 2 = 0 (t−1)(t−2)=0(t-1)(t-2) = 0

t=1 s and t=2 s\boxed{t = 1 \text{ s and } t = 2 \text{ s}}

(c) Position when acceleration is zero:

Set a(t) = 0: 12t−18=012t - 18 = 0 t=1.5 st = 1.5 \text{ s}

x(1.5)=2(1.5)3−9(1.5)2+12(1.5)+5x(1.5) = 2(1.5)^3 - 9(1.5)^2 + 12(1.5) + 5 x(1.5)=6.75−20.25+18+5x(1.5) = 6.75 - 20.25 + 18 + 5

x=9.5 m\boxed{x = 9.5 \text{ m}}

2Problem 2hard

❓ Question:

A particle moves in the xy-plane with position vector r⃗(t)=(3t2)i^+(4t−t3)j^\vec{r}(t) = (3t^2)\hat{i} + (4t - t^3)\hat{j} meters. Find: (a) the velocity and acceleration vectors at t = 2 s, (b) the speed at t = 2 s, and (c) the tangential and normal components of acceleration at this time.

💡 Show Solution

Given: r⃗(t)=3t2i^+(4t−t3)j^\vec{r}(t) = 3t^2\hat{i} + (4t - t^3)\hat{j}

(a) Velocity and acceleration at t = 2 s:

v⃗(t)=dr⃗dt=6ti^+(4−3t2)j^\vec{v}(t) = \frac{d\vec{r}}{dt} = 6t\hat{i} + (4 - 3t^2)\hat{j}

At t = 2: v⃗(2)=12i^+(4−12)j^=12i^−8j^\vec{v}(2) = 12\hat{i} + (4 - 12)\hat{j} = 12\hat{i} - 8\hat{j}

v⃗(2)=12i^−8j^ m/s\boxed{\vec{v}(2) = 12\hat{i} - 8\hat{j} \text{ m/s}}

a⃗(t)=dv⃗dt=6i^−6tj^\vec{a}(t) = \frac{d\vec{v}}{dt} = 6\hat{i} - 6t\hat{j}

a⃗(2)=6i^−12j^ m/s2\boxed{\vec{a}(2) = 6\hat{i} - 12\hat{j} \text{ m/s}^2}

(b) Speed at t = 2 s:

v=∣v⃗∣=122+(−8)2=144+64v = |\vec{v}| = \sqrt{12^2 + (-8)^2} = \sqrt{144 + 64}

v=208=14.4 m/s\boxed{v = \sqrt{208} = 14.4 \text{ m/s}}

(c) Tangential and normal components:

Unit tangent: T^=v⃗∣v⃗∣=12i^−8j^14.4\hat{T} = \frac{\vec{v}}{|\vec{v}|} = \frac{12\hat{i} - 8\hat{j}}{14.4}

Tangential acceleration: aT=a⃗⋅T^=(6)(12)+(−12)(−8)14.4=72+9614.4a_T = \vec{a} \cdot \hat{T} = \frac{(6)(12) + (-12)(-8)}{14.4} = \frac{72 + 96}{14.4}

aT=11.7 m/s2\boxed{a_T = 11.7 \text{ m/s}^2}

Normal acceleration: aN=∣a⃗∣2−aT2a_N = \sqrt{|\vec{a}|^2 - a_T^2}

∣a⃗∣=36+144=13.4 m/s2|\vec{a}| = \sqrt{36 + 144} = 13.4 \text{ m/s}^2

aN=(13.4)2−(11.7)2=179.6−136.9a_N = \sqrt{(13.4)^2 - (11.7)^2} = \sqrt{179.6 - 136.9}

aN=6.5 m/s2\boxed{a_N = 6.5 \text{ m/s}^2}

3Problem 3easy

❓ Question:

An object falls from rest. Its acceleration due to air resistance is given by a = g - bv, where g = 9.8 m/s², b = 0.2 s⁻¹, and v is the velocity. Find: (a) the terminal velocity, (b) the velocity as a function of time, and (c) the time to reach 95% of terminal velocity.

💡 Show Solution

Given:

  • a = g - bv
  • g = 9.8 m/s²
  • b = 0.2 s⁻¹
  • v₀ = 0

(a) Terminal velocity:

At terminal velocity, a = 0: 0=g−bvt0 = g - bv_t

vt=gb=9.80.2v_t = \frac{g}{b} = \frac{9.8}{0.2}

vt=49 m/s\boxed{v_t = 49 \text{ m/s}}

(b) Velocity as function of time:

dvdt=g−bv\frac{dv}{dt} = g - bv

Separating variables: dvg−bv=dt\frac{dv}{g - bv} = dt

−1bln⁡(g−bv)=t+C-\frac{1}{b}\ln(g - bv) = t + C

At t = 0, v = 0: C=−1bln⁡(g)C = -\frac{1}{b}\ln(g)

ln⁡(g−bvg)=−bt\ln\left(\frac{g - bv}{g}\right) = -bt

g−bv=ge−btg - bv = ge^{-bt}

v=gb(1−e−bt)v = \frac{g}{b}(1 - e^{-bt})

v(t)=49(1−e−0.2t) m/s\boxed{v(t) = 49(1 - e^{-0.2t}) \text{ m/s}}

(c) Time to reach 95% of v_t:

0.95vt=vt(1−e−bt)0.95v_t = v_t(1 - e^{-bt}) 0.95=1−e−bt0.95 = 1 - e^{-bt} e−bt=0.05e^{-bt} = 0.05 −bt=ln⁡(0.05)-bt = \ln(0.05)

t=−ln⁡(0.05)b=−−2.9960.2t = -\frac{\ln(0.05)}{b} = -\frac{-2.996}{0.2}

t=15.0 s\boxed{t = 15.0 \text{ s}}

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Derivatives and integrals in kinematics
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