9/26 do now

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 vx = dx/dt to find x-velocity. And we can use ax = dvx/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 ax(t).

4. Let’s first consider a graphical approach. • Suppose a graph of ax-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 t1 and t1 into many smaller intervals, calling a typical one ∆t. During this ∆t, the average acceleration is aav-x. the change in x-velocity ∆vxduring ∙∆t is ∆vx = ∑aav-x∙∆t The total x-velocity change ∆v is represented graphically total area under the ax-t cure between the vertical lines t1 and t2. ∆v

5. ∆vx = ∑aav-x∙∆t • In the limit that all the ∆t’s become very small and the number of ∆t’s become very large, the value of aav-x for the interval from any time t to t + ∆t approaches the instantaneous x-acceleration axat time t. • In this limit, the area under the ax-t curve is the integralof ax (which is in general a function of t) from t1 to t2. If v1xis the x-velocity of body at time t1 and v2x is the velocity at time t2, then

6. We can carry out exactly the same procedure with the curve of vx-t graph. If x1is a body’s position at time t1 and x2 is its position at time t2, we know that ∆x = vav-x∙∆t, where vav-x is the average x-velocity during ∆t. • the total displacement x2 – x1 during the interval t2 – t1 is given by ∆x = ∑vav-x∙∆t ∆x

7. ∫xn dx = + C ∫xndx = - or F’(x) = f(x) If ∫ exdx = eb - ea b b a a bn+1 xn+1 an+1 b ∫ sinx dx = -cosb + cosa n+1 n+1 n+1 a dF(x) = f(x) b b b ∫ ∫ ∫ cosx dx = sinb – sina dx dx = b - a f(x) = F(b) – F(a) a a a ∫ dx = ln x + C 1 1 x x Some simple integrals ∫ dx = x + C ∫ex dx = ex + C ∫sinx dx = -cosx + C ∫cosx dx = sinx + C b ∫ dx = ln b - ln a = ln (b/a) a

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

9. 2 2 • ∫ • ∫ • ∫ x2 dx x-2 dx x3 dx 0 0 • ∫ sinx dx 0 π • ∫ cosx dx 0 b π 8 Practice - evaluate the following 2 • ∫ (3x3 + 2x2 – ½ x + 3)dx 0

12. a. ax = 2.0 m/s2 – (0.10 m/s3)t vx(t) = v0 + (2.0 m/s2)t – (0.10 m/s3)(1/2)t2 v0 = 10 m/s (given) vx(t) = 10 m/s + (2.0 m/s2)t – (0.05 m/s3)t2

13. a. vx(t) = 10 m/s + (2.0 m/s2)t – (0.05 m/s3)t2 x(t) = x0 + (10 m/s)t + (2.0 m/s2)(1/2)t2 – (0.05 m/s3)(1/3)t3 x0 = 50 m (given) x(t) = 50 m + (10 m/s)t + (1.0 m/s2)t2 – (0.05/3 m/s3)t3

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

15. C. We find the maximum x-velocity by substituting t = 20 s (when x-velocity is maximum) into the equation for vx from part (a): vx(t) = 10 m/s + (2.0 m/s2)t – (0.05 m/s3)t2 vx(20 s) = 10 m/s + (2.0 m/s2)(20 s) – (0.05 m/s3)(20 s)2 vx(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/s2)t2 – (0.05/3 m/s3)t3 x(20 s) = 50 m + (10 m/s)(20 s) + (1.0 m/s2)(20 s)2 – (0.05/3 m/s3)(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/s3)t for t >= 0. How far does the object travel in the first 3 seconds? 9 m a = (2 m/s3)t vx = (1 m/s3)t2 vx = vox + (2 m/s3)½ t2 vox = 0 vx = (1 m/s3)t2 x = xo + (1 m/s3)(1/3)t3 x(3 s) – x (0 s) = (1 m/s3)(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 vx-t graph be • A straight line, • Concave up • 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: ∫ eaxdx • Let u = ax, du = a dx, dx = (1/a) du • ∫ eax dx = ∫ eu (1/a) du = (1/a) eu = (1/a)eax

21. ∫ • ∫ • ∫ • ∫ • ∫ x-1 dx sin2x dx 3x2 dx (x-a)2 dx e-2x dx x dx (d2+x2)1/2 Practice - evaluate the following π 0 2 0 8 0 a 0 b a a • ∫ 0

22. Class work – differential and integral