Calculus Review. v = x t

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1 Calculus Review Instructor : Kim 1. Average Rate of Change and Instantaneous Velocity To find the average velocity(v ) of a particle, we need to find the particle s displacement (=change in position) divided by the time interval v = Q1) A particle s motion can be described by the position function x(t) = t² +2t. Find the average velocity between t =2 ~5s. Sol) = x f x i = x(5) x(2) = 35 8 = 9m/s t f t i x(t) x(t)=t 2 +2t 9m/s 35 The linear slope represents the average velocity between the time interval t = 2~5s t(s) However, finding the average velocity is not useful information. We are more interested in finding the velocity or speed at a certain specific time. In order to do that, we can make the time interval shorter. Let s say we want to find the velocity at a specific time t=2s. If we find the average velocity between t=2~2.1s, then we can find the velocity coming close to the actual velocity at t=2s. Example1)The average velocity during t = 2~2.1s Position function is x(t) = t² +2t, so x(t) x(t)=t 2 +2t m/s = x(2.1) x(2) = = 6.1m/s The linear slope represents the average velocity between the time interval t = 2~2.1s t(s)

2 If we make the time interval much shorter and shorter, we will get much closer to the velocity at t=2s. Example2) The average velocity during t = 2~2.0001s Position function : x(t) = t² +2t x(t)=t 2 +2t x(t) m/s = x(2.0001) x(2) = = m/s The linear slope represents the average velocity between the time interval t = 2~2.0001s t(s) From Example2), since = t f t i, = s. Here, represents a very short time interval. Then = x(2+) x(2), where = s The above equation allows us to find velocity that is fairly close to t =2s. If we express an equation to find the velocity at any specific time t = t o, then = x(t o+) x(t o ), where is a very short time interval If we eventually make the so close to zero(= ), then we are actually find the actual velocity at t = t o. Making very very very close to zero is expressed as lim 0 So finding the instantaneous velocity at any specific time t = t o is v =lim 0 x(t o +) x(t o ) : instantaneous velocity at t =t 0 o Q2) A particle s motion can be described by the position function x(t) = t² +2t. Find the instantaneous velocity at t=2s. Method1) Using derivatives v 0 = x(t o+) x(t o ) x(2+) x(2) { (2+Δt) 2 +2(2+Δt)} { } Δt+Δt Δt 8 Δt 2 +6Δt 0 0 (ΔΔt +6)=6m/s 0 Method2) Using x x o =v o t + 1 a 2 xt² and v = v o + a x t x(t) = t² +2t shows that x o =0, v o =2m/s and a x =2m/s 2. So v = v o +a x t = =6m/s

3 v = : average velocity v =lim 0 x(t o +) x(t o ) : instantaneous velocity at t =t 0 o Practice Questions *~Use derivatives only! 1. The position of a particle moving along the x-axis x(t) = 2t 2. What is the average velocity during the time interval t =1s to t=3s? What is the velocity at t =3s? a) 6m/s, 12m/s b) 8m/s, 12m/s c) 6m/s, 14m/s d) 8m/s, 14m/s 2. The position of a particle moving along the x-axis x(t) = 2 + 4t + 3t 2. What is the average velocity during the time interval t =1s to t=3s? a) 4m/s b) 12m/s c) 16m/s d) 22m/s 3. The position of a particle moving along the x-axis x(t) = 2 + 4t + 3t 2. What is the velocity at t=3s? a) 4m/s b) 12m/s c) 16m/s d) 22m/s 4. The position of a particle moving along the x-axis x(t) = t 6t 2. What is the average velocity during the time interval t =1s to t=3s? a) 6m/s b) 4m/s c) 2m/s d) 8m/s

4 v =lim 0 x(t o +) x(t o ) : instantaneous velocity at t =t 0 o 5. The position of a particle moving along the x-axis x(t) = t 6t 2. What is the velocity at t=3s? a) 14m/s b) 12m/s c) 14m/s d) 12m/s 2. Rules for Differentiating Functions If we summarize the derivative for all the position functions in the practice questions, then x(t) = 2t 2 => v 0 = 4t x(t) = 2 + 4t + 3t 2 => v 0 = 4 + 6t x(t) = t 3t 2 => v = 22 6t 0 Following the patterns, we can devise a simple method of finding the derivatives of the position functions. If x = t n, then lim 0 = ntn 1 Also, rather expressing the derivatives as lim 0 dx, a more simple expression is Δx v x = dx Δt 0 : instantaneous speed Δt

5 Solve the following two questions again using v x = dx, where if x = t n then v x = dx n 1 = nt 3. The position of a particle moving along the x-axis x(t) = 2 + 4t + 3t 2. What is the velocity at t=3s? a) 4m/s b) 12m/s c) 16m/s d) 22m/s 5. The position of a particle moving along the x-axis x(t) = t 6t 2. What is the velocity at t=3s? a) 14m/s b) 12m/s c) 14m/s d) 12m/s The following question describes a motion where the acceleration is not constant. So it can only be solved using calculus. 6. The position of a particle moving along the x axis is given by x(t) = 6t 2 t 3. i) Find the position of the particle when t=1s, 2s and 3s ii) Express the velocity function using v x = dx iii) Find the velocity when t=1s, 2s, 3s and 4s

6 iv) Plot a 'velocity vs time' graph below and find the position of the particle when it achieves its maximum speed in the x direction. v(t) x(t) 3. Rules of Integration The difference between squaring and taking the square root is that if we square a positive number and then take the square root of the result, the positive square root value will be the number that you squared. ex) 4 2 =16 and 16 = 4 In the same way, if we can obtain the velocity by differentiating the position or obtain the acceleration by differentiating the velocity, then we can work opposite by obtaining the position by anti-differentiating the velocity or obtain the velocity by anti-differentiating the acceleration. A synonym for anti-differentiation is integration. That is, integration is the inverse of differentiation. Basic Rule If x is any variable, and c is any constant, then 7. Integrate the x x n dx = 1 n+1 xn+1 + c, where n 1 8. Integrate the x 3 9. Integrate the 3x

7 Deriving the position function x from the velocity function v from acceleration function a. Since v = dx, so x = v, Deriving the velocity function v from the acceleration function a. Since a = dv, so v = a x n dx = 1 n+1 xn+1 + c, where n 1 Questions An object moving in a straight line has a velocity v in meters per second that varies with time t in seconds according to the following function. v = t The instantaneous acceleration of the object at t = 2 seconds is (a) 2 m/s 2 (b) 4 m/s 2 (c) 5 m/s 2 (d) 6 m/s 2 (e) 8 m/s If the position of the object was at x = 5m when t = 0s, then find the position of the object at t = 6s. (a) 60m (b) 65 m (c) 70 m (d) 75 m (e) 80 m 12. The displacement of the object between t = 0 and t = 6 seconds is (a) 22 m (b) 28 m (c) 40 m (d) 42 m (e) 60 m 13. A particle moves along the x-axis with a non-constant acceleration described by a = 12t(m/s 2 ). If the particle starts from rest so that its speed v and position x are zero when t = 0, where is it located when t = 2 seconds? (a) x = 12 m (b) x = 16m (c) x = 24 m (d) x = 32 m (e) x = 48 m

Physics 101 Lecture 3 Motion in 1D Dr. Ali ÖVGÜN

Physics 101 Lecture 3 Motion in 1D Dr. Ali ÖVGÜN EMU Physics Department Motion along a straight line q Motion q Position and displacement q Average velocity and average speed q Instantaneous velocity and