1. 9/26 do now Quick quiz

2. Homework #4 – due Friday, 9/30 Reading assignment: 2.1 – 2.6; 3.1-3.2 Questions: 2.67, 72, 76, 88 – the solutions are on the school website. Homework – due Tuesday, 10/4 – 11:00 pm Mastering physics wk 4

3. Objective – velocity and position by integration In the case of straight-line motion, if the position x is a known function of time, we can find v x = dx/dt to find x-velocity. And we can use a x = dv x /dt to find the x- acceleration as a function of time In many situations, we can also find the position and velocity as function of time if we are given function a x (t).

4. Let’s first consider a graphical approach. Suppose a graph of a x -t is given as shown in the graph. We learned that the area under the graph represents the change in velocity. To calculate the area, we can divide the time interval between times t 1 and t 1 into many smaller intervals, calling a typical one ∆t. During this ∆t, the average acceleration is a av-x. the change in x-velocity ∆v x during ∙∆t is ∆v x = ∑a av-x ∙∆t The total x-velocity change ∆v is represented graphically total area under the ax-t cure between the vertical lines t 1 and t 2. ∆v

5. In the limit that all the ∆t’s become very small and the number of ∆t’s become very large, the value of a av-x for the interval from any time t to t + ∆t approaches the instantaneous x-acceleration a x at time t. In this limit, the area under the a x -t curve is the integral of a x (which is in general a function of t) from t 1 to t 2. If v 1x is the x-velocity of body at time t 1 and v 2x is the velocity at time t 2, then ∆v x = ∑a av-x ∙∆t

6. We can carry out exactly the same procedure with the curve of v x -t graph. If x 1 is a body’s position at time t 1 and x 2 is its position at time t 2, we know that ∆x = v av-x ∙∆t, where v av-x is the average x-velocity during ∆t. the total displacement x 2 – x 1 during the interval t 2 – t 1 is given by ∆x = ∑v av-x ∙∆t ∆x

7. Some simple integrals ∫ dx = x + C ∫ x n dx = + C x n+1 n+1 ∫ e x dx = e x + C ∫ a b dx = b - a ∫ ab f(x) = F(b) – F(a) or F’(x) = f(x) dF(x)dF(x) dxdx = f(x) If ∫ x n dx = - b n+1 n+1 a b a n+1 n+1 ∫ sinx dx = -cosx + C ∫ cosx dx = sinx + C a ∫ sinx dx = -cosb + cosa b ∫ e x dx = e b - e a b a ∫ cosx dx = sinb – sina a b ∫ dx = ln x + C x 1 ∫ dx = ln b - ln a = ln (b/a) x 1 b a

8. More properties of integrals ∫ (u + v) dx = ∫ u dx + ∫ v dx ∫ au dx = a ∫ u dx

9. Practice - evaluate the following 1. ∫ x 2 dx 0 2 3. ∫ x -2 dx 8 b 2. ∫ x 3 dx 0 2 6. ∫ cosx dx π 0 5. ∫ sinx dx 0 π 4. ∫ (3x 3 + 2x 2 – ½ x + 3) dx 0 2

12. a. a x = 2.0 m/s 2 – (0.10 m/s 3 )t v x (t) = v 0 + (2.0 m/s 2 )t – (0.10 m/s 3 )(1/2)t 2 v 0 = 10 m/s (given) v x (t) = 10 m/s + (2.0 m/s 2 )t – (0.05 m/s 3 )t 2

13. a. x(t) = x 0 + (10 m/s)t + (2.0 m/s 2 )(1/2)t 2 – (0.05 m/s 3 )(1/3)t 3 x 0 = 50 m (given) v x (t) = 10 m/s + (2.0 m/s 2 )t – (0.05 m/s 3 )t 2 x(t) = 50 m + (10 m/s)t + (1.0 m/s 2 )t 2 – (0.05/3 m/s 3 )t 3

14. b. The maximum value of v x occurs when the x- velocity stops increasing and begins to decrease. At this instant, dv x /dt = a x = 0.

15. C. We find the maximum x-velocity by substituting t = 20 s (when x-velocity is maximum) into the equation for v x from part (a): v x (t) = 10 m/s + (2.0 m/s 2 )t – (0.05 m/s 3 )t 2 v x (20 s) = 10 m/s + (2.0 m/s 2 )(20 s) – (0.05 m/s 3 )(20 s) 2 v x (20 s) = 30 m/s

16. d. To obtain the position of the car at that time, we substitute t = 20 s into the expression for x from part (a): x(t) = 50 m + (10 m/s)t + (1.0 m/s 2 )t 2 – (0.05/3 m/s 3 )t 3 x(20 s) = 50 m + (10 m/s)(20 s) + (1.0 m/s 2 )(20 s) 2 – (0.05/3 m/s 3 )(20 s) 3 x(20 s) = 517 m

18. example An object initially at rest experiences a time-varying acceleration given by a = (2 m/s 3 )t for t >= 0. How far does the object travel in the first 3 seconds? 9 m a = (2 m/s 3 )t v x = v ox + (2 m/s 3 )½ t 2 v ox = 0 v x = (1 m/s 3 )t 2 x = x o + (1 m/s 3 )(1/3)t 3 x(3 s) – x (0 s) = (1 m/s 3 )(1/3)(3 s) 3 x(3 s) – x (0 s) = 9 m

19. Check your understanding 2.6 If the x-acceleration ax is increasing with time, will the v x -t graph be 1.A straight line, 2.Concave up 3.Concave down

20. Integral by substitution To find the integral of a function that is more complicated than the simple function, we use a technique which is integral by substitution (aka u sub). For example: ∫ e ax dx Let u = ax, du = a dx, dx = (1/a) du ∫ e ax dx = ∫ e u (1/a) du = (1/a) e u = (1/a)e ax

21. Practice - evaluate the following 0 π 1. ∫ sin2x dx 2. ∫ 3x 2 dx 0 2 3. ∫ e -2x dx 0 8 4. ∫ (x-a) 2 dx 0 a 6. ∫ 0 a 5. ∫ x -1 dx a b (d 2 +x 2 ) 1/2 x dx

22. Class work – differential and integral