9/26 do now

• Quick quiz

Homework #4 – due Friday, 9/30Reading 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

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).

• 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 ∆vx during ∙∆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

• 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 ax at time t.

• In this limit, the area under the ax-t curve is the integral of ax (which is in general a function of t) from t1 to t2. If v1x is the x-velocity of body at time t1 and v2x is the velocity at time t2, then

∆vx = ∑aav-x∙∆t

• We can carry out exactly the same procedure with the curve of vx-t graph. If x1 is 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

Some simple integrals

∫ dx = x + C

∫xn dx = + Cxn+1

n+1

∫ex dx = ex + C

∫a

b

dx = b - a

∫a

b

f(x) = F(b) – F(a)or F’(x) = f(x)dF(x)

dx= f(x)If

∫xndx = -bn+1

n+1a

b an+1

n+1

∫sinx dx = -cosx + C

∫cosx dx = sinx + C

a∫ sinx dx = -cosb + cosa

b

∫ exdx = eb - ea b

a

∫ cosx dx = sinb – sina a

b

∫ dx = ln x + Cx1

∫ dx = ln b - ln a = ln (b/a)x1b

a

More properties of integrals

∫ (u + v) dx = ∫ u dx + ∫ v dx

∫ au dx = a ∫ u dx

Practice - evaluate the following

1. ∫ x2 dx0

2

3. ∫ x-2 dx8

b

2. ∫ x3 dx0

2

6. ∫ cosx dxπ

0

5. ∫ sinx dx0

π

4. ∫ (3x3 + 2x2 – ½ x + 3) dx0

2

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

a.

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)

vx(t) = 10 m/s + (2.0 m/s2)t – (0.05 m/s3)t2

x(t) = 50 m + (10 m/s)t + (1.0 m/s2)t2 – (0.05/3 m/s3)t3

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

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

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

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 = vox + (2 m/s3)½ t2

vox = 0

vx = (1 m/s3)t2

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

Check your understanding 2.6• If the x-acceleration ax is increasing with

time, will the vx-t graph be

1. A straight line,

2. Concave up

3. Concave down

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

Practice - evaluate the following

0

π

1. ∫ sin2x dx

2. ∫ 3x2 dx0

2

3. ∫ e-2x dx0

8

4. ∫ (x-a)2 dx0

a

6. ∫0

a

5. ∫ x-1 dxa

b

(d2+x2)1/2

x dx

• Class work – differential and integral