Applied-Physics.pdf

# Chapter 01: Intro to units, errors and graphs ## Physical Quantities The quantities that are measurable are called physical quantities. ## Representation Categories | Category | MKS | CGS | FPS | |---|---|---|---| | Mass | kg | gm | pounds | | Time | sec | sec | sec | | Length | m | cm | ft | #...

# Chapter 01: Intro to units, errors and graphs ## Physical Quantities The quantities that are measurable are called physical quantities. ## Representation Categories | Category | MKS | CGS | FPS | |---|---|---|---| | Mass | kg | gm | pounds | | Time | sec | sec | sec | | Length | m | cm | ft | ## Q1: Convert the following quantities in ft/ sec: a) 1 mile/ hr Sol: 1 mile = 5,280 ft, 1 hour = 3,600 sec 5280 / 3600 = 1.466 ft/ sec b) 28 mile/hr Sol: 28 x 5280 / 3600 = 41.066 ft/ sec ## Q2: Convert 25 ft/ sec to mile/ hr: Sol: 1 mile = 5280 ft, 1 ft = 1/5280 miles, 1 sec = 1/3600 hour 25/5280 miles = 25 x 3600 / 5280 = 17.04 mile/ hr # Errors: The difference between measured and true value is called error. There are two types of errors: ### 1. Random Error: Random error is occurred when repeated measurement of a quantity gives different values under the same condition. It is caused due to unknown reason. ### 2- Systematic Error: Systematic error refers to an effect that influence all the measurements of a particular physical quantity, equally it produces a consistense difference in measurements. Examples are: * Instrumental error * Calculation error # Terminologies: ### Accuracy: A measure of how close the observed value is to the true/actual/ideal value. ### Deviation: A measure of how far the observed value is to the true value. Deviation is mathematical representation of accuracy. $S_i = \frac{\sum_{i=1}^{n} X_i - X}{n}$ x - measurement of all individual. n - number of individuals ### Average deviation: Average of all the deviation values. $\frac{\sum_{i=1}^{n} z_n}{n}$ ### Standard deviation: Standard deviation (ơ) is defined as square root of mean square deviation for an infinite set. $σ = \sqrt{\frac{\sum_{i=1}^{n}(x_i-x)^2}{n-1}} = \sqrt{\frac{\sum_{i=1}^{n} data\_i^2}{n-1} - avg. of all data}^2$ ### Standard Error: Standard errors are caused due to standard deviation. Standard Error = σ / √n ### Probable Error: ± 0.645 x Standard error, where ±0.645 is natural constant. # Combining effects of errors: There are three possibilities: ### 1. Error in Sum: Let measurement of one quantity be a ± Sa and measurement of another quantity be b ± Sb. Then their addition will be: (a ± Sa) + (b ± Sb) = (a + b) ± (Sa + Sb) ## Q: Find the equivalent resistance in series where R₁ = 100 ± 2Ω & R₂ = 200 ± 3Ω Sol: $$R_E=R_1+R_2=(100 ± 2) + (200 ± 3)$$ $$= (100 + 200) ± (2 + 3)$$ $$= 300 ± 5Ω$$ ### 2. Error in Difference: (a ± Sa) - (b ± Sb) = (a-b) ± (Sa + Sb) ### 3. Error in Multiplication/Division: If two quantities are multiplied / divided then error in them is found by $S_Q = ± \frac{(Sa + Sb)}{a+b}$ → Standard err: in 'a' Standard err: in 'b' Value of multiplication/division of error free part ## Q: Capacitance of capacitor is 2 ± 0.1 F, voltage is 25 ± 0.5 V. Find Charge (Q)? Sol: Q = CV = 2 x 25 = 50C Now for finding error: $S_q = ±(S_c + S_v)= (+ \frac{0.1}{2}) + (+\frac{0.5}{25})$ $S_c = \frac{0.1}{Q} = 0.07$, $S_q = 50 x 0.07$ = 3.5 Q = Q ± SQ = 50 ± 3.5 46.5 ≤ Q ≤ 33.5 ## Q : The mass of an object is 345.1 ± 0.1 gm and volume is 41.55 ± 0.05 cm³. Find density: Sol: D = M/V = 345.1 / 41.55 = 8.3 Now for finding error: $S_D = ±(\frac{S_m}{m} + \frac{S_v}{V}) = ±( \frac{0.1}{345.1} + \frac{0.05}{41.55})$ $S_p = ± 0.00149$, $S_p = ± 8.3 x 0.00149 = 0.0123$ D=D+SD = 8.3±0.0123 g/ cm³ 8.2877 ≤ D ≤ 8.3123 # Graphs: A diagram showing the relation between two variables. ## Rules of Graph: * Find dependent and independent variable * Decide Scale. * Define starting point. * Plot atleast 6 values. ## Shapes of Graphs: ### a) Linear/straight Line eg: $y=mx+c$ * y - intercept * m - slope ### b) Exponential / Power equations: * $y = ke^x$ * $y = ke^{-x}$ * $y = kx^2$ # Straight Line transformation: | f(x) | dependent(y) | independent (x) | Slope | Intercept | |---|---|---|---|---| | y = mx + c | y | x | m | c | | y = bx + a | y | x | b | a | | y = b√x + c | y | √x | b | 1/c | | y = ac^x | logy | x | loga | loga | # Types of Graphs: ### 1) Linear Graph: ### 2) Semi-log Graph: ### 3) log-log graph: # Numerical: Fit the given data to a straight line graph and calculate from the given observations. | length (cm) | time (sec) | |---|---| | 62.0 | 1.58 | | 72.0 | 1.71 | | 85.0 | 1.85 | | 100.0 | 2.01 | | 112.0 | 2.12 | | 120.0 | 2.20 | | 134.0 | 2.32 | | 148.0 | 2.44 | ### a) Standard deviation First of find all the values of 'g' using the formula $g = \frac{2}{4\iota\iota}$ by putting the data given above one by one. * g₁ = 980.47, g₂ = 972.07, g₃ = 980.47, g₄ = 977.16, g₅ = 983.79, g₆ = 978.80, g₇ = 982.85, g₈ = 981.39 * g_avg = 979.63 Standard deviation = $σ= \sqrt{\frac{\sum_{i=1}^{n} g_n^2}{n-1}} = σ= \sqrt{\frac{(g_1 - g)^2 + (g_2 - g) + ... + (g_8 - g)^2}{7}} ≈ 3.70$ ### b) Standard error: Standard error = $σ / \sqrt{n} = 3.70 / \sqrt{8} ≈ 1.30$ ### c) Probable error: Probable error = ±0.6745 x Standard error = ±0.6745 x 1.30 = ±0.876 ≈ ±0.88 ### d) Final value: g = 979.63 ± 0.88 cm/s² 978.75 ≤ g ≤ 980.51 # Chapter 02: Vector Analysis ## Scalar quantities: Magnitude + Unit ## Vector quantities: Magnitude + Unit + Direction ## Co-ordinate systems: ### 1) Rectangular/cartesian : ### 2) Cylindrical: * (x, y, z) * (ρ, θ, z) * (x, α) * (ρ, θ, θ) * (R, θ, φ) ### 3) Spherical: * Polar angle θ * azimuthal angle φ ## Transformations: ### 1) Rect: to Cylindrical (x, y, z) ⇒ (ρ, θ, z) * ρ = √x² + y² * θ = tan-¹(y/z) * z = z ### 2) Cylindrical to Rectangular (ρ, θ, z) ⇒ (x, y, z) * x = ρ cos θ * y = ρ sin θ * z = z ### 3) Rectangular to spherical (R, θ, φ) ← (x, y, z) * R = √x² + y² + z² * θ = tan-¹(y/x) * φ = tan-¹(√x²+y²/z) ### 4) Spherical to rectangular (R, θ, φ) = (x, y, z) * x = R sin θ cos φ * y = R sin θ sin φ * z = R cos φ ### 5) Cylindrical to spherical (ρ, θ, z) ⇒ (R, θ, φ) * R = √ρ² + z² * φ = tan-¹(ρ/z) * θ= θ ### 6) Spherical to cylindrical (R, θ, φ) ⇒ (ρ, θ, z) ## Q: Convert the following by using transformations: ### a) (2, 4π/3, 8) to rectangular: * x = ρ cos θ = 2 x cos (4π/3) = -1 * y = ρ sin θ = 2 x sin (4π/3) = -√3 * z = z = 8 ⇒ (-1, √3, 8) ### b) (3, π/4, π/3) into rectangular: * x = R sin θ cos φ = 3 x sin(π/4) x cos(π/3) = - 3√2/4 * y = R sin θ sin φ = 3 x sin (π/4) x sin(π/3) = - 3√6/4 * z = R cos φ = 3 x cos(π/3) = -3/2 ⇒(-3√2/4, - 3√6/4, -3/2) ### c) (2, 30°, 4) into spherical: * R = √ρ² +z² = √2² + 4² = √20 = 2√5 * θ = ρ = 30° = π/6 * φ = tan-¹(ρ/z) = tan-¹(2/4) = 26.5 ⇒ (2√5, π/6, 26.5) ### d) (√3, 1, 4) to cylindrical: * ρ = √x² + y² = √(√3)² + 1² = √3 + 1 = √4 = 2 * θ = tan-¹(y/x) = tan-¹(1/√3) = 30° * z = z = 4 ⇒ (2, 30°, 4) # Differentiations The process of finding the function that outputs the rate of change of one variable with respect to another. If function is vector quantity then its differentiation is called vector differentials. ## Q: Given function: R = sin(t) + cos(t)j + k Find: a) dR/dt b) d²R/dt² c) |dR/dt| d) |d²R/dt²| a) R = sin(t)i + cos(t)j + k dR/dt = cos(t)i - sin(t)j + k ------ (i) b) Differentiating (i) again, we get: d²R/dt² = -sin(t)i - cos(t)j ------ (ii) c) Taking magnitude of (i), we get: |dR/dt| = √(cos(t))² + (-sin(t))² + (1)² = √cos²t + sin²t + 1 |dR/dt| = √2 d) Taking magnitude of (ii), we get: |d²R/dt²| = √(sin(t))² + (-cos(t))² = √sin²t + cos²t |d²R/dt²| = 1 ## Q: x = e^-t, y = 2cos3t , z = 2sin3t, where t is time. Find: a) Velocity and acceleration at time 't' b) Magnitude of velocity and acceleration at time t = 0 Sol: f(t) = x + y + z f(t) = e^-t + 2cos3t + 2sin 3t ### a) For Finding velocity differentiate f(t) f '(t) = -e^-t - 6sin(t) + 6 cos(3t) --- (i) For find acceleration differentiate (i) f ''(t) = e^-t - 18cos3t - 18sin(3t) --- (ii) ### b) For velocity, put t = 0 in (i) and find its magnitude: velocity = √(e^-0)² + (-6sin(0))² + (6cos(3(0)))² = √(1 + 0 + 6²) = √37 For acc:, put t = 0 in (ii) and find its magnitude. acc: = √(e^-0)² + (-18cos(3(0)))² + (-18sin(3(0))² = √(1 + 18² + 0) acc: = √1+ 324 ≈ 18.02 # Gradient of a vector: Let Φ(x, y, z) be defined and differentiable at each point (x, y, z) in a certain region of space (i.e. Φ defines a differentiable scalar field). Then the gradient of Φ, written ∇Φ or grad Φ, is defined by: ∇Φ = $\frac{δΦ}{δx}$i + $\frac{δΦ}{δy}$j + $\frac{δΦ}{δz}$k = $\frac{δΦ}{δx}$i + $\frac{δΦ}{δy}$j + $\frac{δΦ}{δz}$k Note that Φ defines a vector field. Kind of used to convert scalar field to vector field. # Directional derivative: The directional derivative is the rate of change at which the function changes at a point in a particular direction. Directional derivative helps in finding slope in 3D-plane. It is denoted by (∇Φ). â ; â = a/|a| ## Q: If Φ(x, y, z) = 3xy - yz², then find: a) ∇Φ b) (∇Φ). â in direction of a = 2i + 3j - k a) ∇Φ = ( $\frac{δ}{δx}$, $\frac{δ}{δy}$, $\frac{δ}{δz}$ )(3xy - yz²) = ( $\frac{δ(3xy-yz^2)}{δx}$i + $\frac{δ(3xy-yz^2)}{δy}$j + $\frac{δ(3xy-yz^2)}{δz}$k) = 6xy²i + (3x - 3yz²)j - 2y²zk ----- (i) △∇Φ(1, -2, -1) = 6(1)(-2) + (3(1)²-3(-2)(-1))j - 2(-2)(-1) = -12i - 9j - 16k ------- (ii) b) (∇Φ). â ; for â = 2i + 3j - 2k = $\frac{2i + 3j - 2k}{\sqrt{2² + 3² + (-2)²}}$ (∇Φ). â = (-12i - 9j - 16k). ($\frac{2i + 3j - 2k}{\sqrt{17}}$) (∇Φ). â = $\frac{-24 - 27 + 32}{\sqrt{17}}$ = $\frac{-19}{\sqrt{17}}$ # Divergence: Divergence is defined as variation in amount of vector quantity. Let D(x, y, z) = V₁i + V₂j + V₃k be defined and differentiable at each point (x, y, z) in a certain region of space (i.e. D defines a differentiable vector field). ∇.D = (i + j + k). (V₁i + V₂j + V₃k) = $\frac{δV_1}{δx} + \frac{δV_2}{δy} + \frac{δV_3}{δz}$ * If ∇.V = 0; then field is Solenoidal. * If ∇. V ≠ 0; then field is non-solenoidal. * Input in divergence is vector but output is scalar. ## Que: Determine the value of 'a' so that the vector field is solenoidal. V = (x + 3y)i + (y - 2z)j + (x + az)k Sol: As we know that vector field is solenoidal when ∇.V = ∇Div = 0 So: ∇.V = 0 (i + j + k). ((x + 3y)i + (y - 2z)j + (x + az)k) = 0 $\frac{δ(x + 3y)}{δx}$ + $\frac{δ(y - 2z)}{δy}$ + $\frac{δ(x + az)}{δz}$ = 0 1 + 1 + a = 0 a = -2 # Curl of a vector: If V(x, y, z) is a differentiable vector field than ∇ x V, curl or rotation of V, written ∇ x V or rot V, is defined by: ∇ x V = (i + j + k) x (V₁i + V₂j + V₃k) = ( $\frac{δ}{δx}$, $\frac{δ}{δy}$, $\frac{δ}{δz}$ ) x (V₁, V₂, V₃) - Curl of vector field is defined as the rotation b/w two quantities in such a way one quantity is moving linearly and other quantity is moving in circular direction. Note: ∇ x V = 0 irrotational field; ∇ x V ≠ 0 rotational or solenoidal field. ## Questions: ### 1. If A = xy²i + 2xyzj + zyz k, Find ∇ x A. Sol: ∇ x A = (i j k) \begin{vmatrix} δ/δx & δ/δy & δ/δz\\ xy² & 2xyz & zyz \end{vmatrix} ∇ x A = { ( $\frac{δ (zyz)}{δy}$ - $\frac{δ (2xyz)}{δz}$ )i - ( $\frac{δ (zyz)}{δx}$ - $\frac{δ (xy²)}{δz}$ )j + ( $\frac{δ (2xyz)}{δx}$ - $\frac{δ (xy²)}{δy}$ )k} = { (2xz) - (2xy) }i - (0 - 0)j + ( (4xyz) - 3xy)k ∇ x A = (2xz - 2xy)i + (4xyz - 3xy)k ### 2. Find curl, if the given field A = (2x² + y)i + (y² + 4xz)j + (3y)k at (2, 1, 1). Sol: ∇ x A = (i j k) \begin{vmatrix} δ/δx & δ/δy & δ/δz\\ 2x² + y & y² + 4xz & 3y \end{vmatrix} ∇ x A = { ( $\frac{δ (3y)}{δy}$ - $\frac{δ (y² + 4xz)}{δz}$ )i - ( $\frac{δ (3y)}{δx}$ - $\frac{δ (2x² + y)}{δz}$)j + ( $\frac{δ (y² + 4xz)}{δx}$ - $\frac{δ (2x² + y)}{δy}$ )k } = { (3 - 4x)}i - (0 - 0)j + (4z - 1)k ---- (v） Putting (2, 1, 1) in (v) ∇ x A = (3 - 4(2))i + (4(1)-1)k ∇ x A = (-5 + 3)k # Date: = ∇F₁ + ∇F₂ + ∇G₁ + ∇F₂ + ∇F₃ + ∇G₃ = ( $\frac{δF_1}{δx} + \frac{δF_2}{δy} + \frac{δF_3}{δz}$) + (- $\frac{δG_1}{δx} + \frac{δG_2}{δy} + \frac{δG_3}{δz}$ ) = ∇F + ∇G ∴ ∇Φ = $\frac{δΦ_1}{δx} + \frac{δΦ_2}{δy} + \frac{δΦ_3}{δz}$ # Summary: | Category | Type | |---|---| | Gradient | Φ → Scalar | | Divergence | V → Vector | | Curl | V → Vector | | | ∇Φ → Vector | | | ∇.V → Scalar | | | ∇ x V → Vector | # Integration: Reverse process of differentiation is called integration. Integration of vector function is called vector integration. ## 3- If A= xyi - zxzj + 2yzF, Find ∇ x (∇ x A) Sol: Find ∇ x A first, then take curl of result. ∇ x A = (i j k) \begin{vmatrix} δ/δx & δ/δy & δ/δz\\ xy & -zxz & 2yz \end{vmatrix} ∇ x A = { ( $\frac{δ (2yz)}{δy}$ - $\frac{δ (-zxz)}{δz}$ )i - ( $\frac{δ (2yz)}{δx}$ - $\frac{δ (xy)}{δz}$ )j + ( $\frac{δ (-zxz)}{δx}$ - $\frac{δ (xy}{δy}$ )k } = {(2xz) - ( -2xz)}i - (0 - 0)j + (-2z - x)k ∇ x A = (2xz + 2x) i - ( -2z - x)k ----- (i) Take curl of (i) ∇ x (∇ x A) = (i j k) \begin{vmatrix} δ/δx & δ/δy & δ/δz\\ 2xz +2x & 0 & -2z - x \end{vmatrix} ∇ x (∇x A) = { ( $\frac{δ (-2z - x)}{δy}$ - $\frac{δ(0)}{δz}$ )i - ( $\frac{δ (-2z - x)}{δx}$ - $\frac{δ (2xz + 2x)}{δz}$ )j + ( $\frac{δ (0)}{δx}$ - $\frac{δ (2xz + 2x)}{δy}$ )k} = (0 - 0) i - (-2 - 2)j + (0 - 0)k ∇ x (∇ x A) = -(-2 - 2)j = 4j ## 4- Prove that: ∇(F + G) = ∇F + ∇G Sol: Let F = F₁i + F₂j + F₃k & G = G₁i + G₂j + G₃k L.H.S ∇(F + G) = ∇ (F₁ + F₂ + F₃ + G₁ + G₂ + G₃) = (i + j + k). (F₁ + F₂ + F₃ + G₁ + G₂ + G₃) = . . . . . ## Q: Let R(t) = (t - t²)i + (2t²)j + (-3)k. Find ∫ R(dt) Sol: ∫ R(t) = ∫(t - t²)i + ∫(2t²)j + ∫(-3)k ∫R(t)dt = ∫t dt - ∫t² dt + ∫2t² dt j - 3∫dt k ∫R(t)dt = $\frac{t²}{2} - ∫t² dt i + \frac{2kt³}{3}$ j - 3t k ∫R(t)dt = ($\frac{t²}{2} - \frac{t³}{3} )i + (\frac{2t³}{3} - 3(2) - (-3)(0))^{-1}$ ∫R(t)dt = ( $\frac{2}{2} - \frac{8}{3}$ )i -( $\frac{3}{3} - \frac{1}{3}$ )j + ( $\frac{16}{2} - \frac{3}{2}$ )j - 3k ∫R(t)dt = (-2 / 3)i - (8/3)j + (13/2)j - 3k [∫R(t)dt]² = (9i - 14i + 15j - 3k ) [∫R(t)dt]³ = (-5i + 15j - 3k **Ans** # Line Integrals: Deals with a line. Let r(u) = x(u)i + y(u)j + z(u)k, where x(u) is the position vector of (x, y, z), defines a curve c joining points P₁ and P₂, where u = u₁ and u = u₂ respectively We assume that c is composed of finite number of curves for each of which x(u) has continuous derivative. Let A(x, y, z ) = A₁i + A₂j + A₃k be a vector function of position defined and continuous along c. Then the integral of the tangential component of A along c from P₁ to P₂, written as ∫ A.dr = ∫A.dr = ∫(A₁dx + A₂dy + A₃dz) C is example of line integral. If A is Force, then line integral defines work done. W = ∫F.dr ## Quesion: Find out the total work done in a force field F = 3xyi - 5zj + 10xk, along the curve x = t² + 1, y = 2t², z = t² from t₁ = 1 to t₂ = 2. Sol: dx = t dt, dy = 4t dt, dz = 2t dt. We have to convert everything in terms of t. W = ∫F.dr = ∫F₁dx + F₂dy + F₃dz W = ∫3xy. 2tdt - ∫5z. 4tdt + ∫10x.2tdt ---- (i) Putting values of x, y and z in ( i) we get W = ∫3(t² +1)(2t²)(2t)dt - ∫20t.t²dt + ∫20t (t² + 1)dt W = ∫(12t⁴ + 12t²)dt - ∫20t³dt + ∫(20t⁴ + 20t²) dt W = ∫12t⁴dt + ∫12t²dt - ∫20t³dt + ∫20t⁴dt + ∫20t²dt = $\frac{12t5}{5}$ + $\frac{12t³}{3}$ + $\frac{20t⁴}{4}$ + $\frac{20t⁵}{5}$ + $\frac{20t³}{3}$ = $\frac{12t⁵}{5}$ + $\frac{12t³}{3}$ + $\frac{20t⁴}{4}$ + $\frac{20t⁵}{5}$ + $\frac{20t³}{3}$ W = 26/3 + 3t⁴ + 12t⁵ + 10t³ W = {2(2⁵) - 2(1⁵)} + {3(2⁴) - 3(1⁴)} + {10(2²) - 10(1²)} W = 126 + 45 + 30 W = 201 Joules **Ans** ## Question: If A = (3x + 6y)i - 14yzj + 20zk. Find ∫A.dr from (0, 0, 0) to (1, 1, 1) along the following path, x = t, y = t², and z = t³. Sol: dx = dt, dy = 2tdt, dz = 3t²dt. ∫¹A.dr = ∫¹A₁dx + A₂dy + A₃dz = ∫¹(3x + 6y)dt + ∫¹(yz) (2tdt) + ∫¹20z(3t²dt) = ∫¹(3t + 6t²)dt - ∫¹t^3(2tdt) + ∫¹20t³(3t²dt) = ∫¹(3t + 6t²)dt - ∫¹2t⁴dt + ∫¹60t⁵dt = [3t²/2 + 2t³]¹₀ - [2t⁵/5]¹₀ + [60t⁶/6]¹₀ = $\frac{3}{2}$ - $\frac{2}{5}$ + 10 - $\frac{3}{2}$ + $\frac{2}{5}$ + $\frac{60}{81}$ = 3t - 4t⁵ + 15 t⁶ = 3 - 4 + 15 = 15 - 1 = 6.5 **Ans** ## Question: The given curve is y = 2x + x². Find ∫r.(xy) that follows (0, 0) - (2, 0) only. Sol: ∫r.(xy) ⇒ First change xy in terms of x by putting y = 2x + x² ∫r.(xy) = ∫r . x(2x + x²) = ∫r . (2x² + x³) = ∫(2x² + x³) dx + (2x² + x³) dy + (2x² + x³) dz = ∫( 6x² + 4x³) dx + 0. dy + 0. dz ∫r.(xy) = ∫( 6x² + 4x³) dx ∫r.(xy) = ∫ 6x² dx + ∫ 4x³ dx ∫r.(xy) = $\frac{6x³}{3}$ + $\frac{4x⁴}{4}$ ∫r.(xy) = 2x³ + x⁴ ∫r.(xy) = 2(2)³ + (2)⁴ ∫r.(xy) = 16 + 16 ∫r.(xy) = 32 **Ans** # Chapter #03 Mechanics Study of motion is called mechanics. ## Cases of motion: * **Resting position (No motion at all)**: constant velocity/motion. Slope = 0 * **Constant Velocity**: constant acceleration. Variable velocity. Slope = variable. * **Constant decreasing velocity**: Deceleration ## Instantaneous Velocity: A Velocity of an object in motion at specific point in time. $V_i = lim┬(Δt→0) Δx/Δt$ ## Instantaneous Acceleration: An acceleration of an object at any specific time during the motion. $a_i = lim┬(Δt→0) ΔV/Δt$ ## Q: Derive the eq: of x(t) by following graph: x(t) - ... Solution: * Δx = x - x₀ = x - x₀ * $V_{avg} = \frac{Δx}{Δt}$ = $\frac{x - x₀}{t - t₀}$ * $V_{avg} = \frac{x - x₀}{t}$ * $V_{avg}t = x - x₀$ * x = $V_{avg}t$ + x₀ ---- ( i ) * $V_{avg}=\frac{ΔV}{Δt} = \frac{V - V₀}{t - t₀}$ * $V_{avg}=\frac{V - V₀}{t}$ * $V_{avg}t = V - V₀$ * V = $V_{avg}t$ + V₀ ---- ( ii) * V avg = V + V₀ / 2 * V = 2$V_{avg}$ - V₀ ---- ( iii ) $V_{avg}t$ + V₀ = 2$V_{avg}$ - V₀ * 2$V_{avg}$ = $V_{avg}t$ + 2V₀ * $V_{avg}$ = V₀ + 1/2$V_{avg}$t ---- ( iv) Replace ( iv ) in ( i) * x = (V₀ + 1/