An=a1+d(n-1)
a1=first term
an=nth term
d=common differnce or how much it increases by each time
n=which one

so
originally 3 trees
a1=3

and after 19 years, 84 trees, so a19=84

a19=84=3+d(19-1)
84=3+d(18)
minus 3 both sides
81=18d
divide both sides by 18
4.5=d

an=3+4.5(n-1) or
an=4.5n-1.5
or
t(n)=3+4.5(n-1) or
t(n)=4.5n-1.5

4 0

3 years ago

A baseball leaves the hand of a pitcher 6 vertical feet above home plate and 60 feet from home plate. Assume the x-axis is a str

d1i1m1o1n [39]

Step-by-step explanation:

Throughout the solution we will use seconds for time, feet for distance, ft/sec for speed, and ft/sec2 for acceleration.

We are assuming that r(0) = (0, 0, 6), v(0) = (130, 0, −3), and a(t) = (0, 0, −32).

Note: These are all vectors in i,j and k directions.

We need to work from a(t), to v(t), to r(t) by integration.

Integrate a(t) to find v(t)

$v(t) = \int\limits {a(t) } \, dt = \int\limits {(0,0,-32)} \, dt = (0,0,-32) *t + C\\$

Using v(0) = (130, 0, −3), we can determine C

C = (130, 0, −3)

Hence, v(t) = (130, 0, 32*t -3) ...... Eq1

Integrate v(t) to find r(t)

$r(t) = \int\limits {(130,0,-3-32*t)} \, dt = (130t , 0 , -3t -16t^2) + C\\$

Using r(0) = (0, 0, 6), we can determine C

C = (0, 0, 6)

Hence, r(t) = (130t, 0, -3t -16t^2 +6) ...... Eq2

Solution

The ball crosses home plate when the x-position is 60 feet using Eq 2

130 t = 60

t = 0.462 seconds

@ t = 0.462 the z-coordinate will represent the height of the ball above the ground

$-3 (0.462) -16 (0.462)^2 +6=1.207 ft$

If we rather wanted this height to be 3 feet when the ball crosses the plate then the vertical component of v(0) will need to be something other that −3.

If we repeat the calculation of r(t) above with v(0) = (130,0,k) then we will find that r(t) = (130t , 0 , 6 + kt - 16t^2)

The ball will still cross home plate at t = 6/13 sec and the height of the ball as it crosses the plate will be :

$k (0.462) -16 (0.462)^2 +6=3 ft\\\\k = \frac{23}{26}$

d) In this case we are assuming that a(t) = (0, 8, −32). If we repeat the calculation of r(t) above with a(t) = (0, 8, −32) and v(0) = v(0) = (130, 0, −3), the only change will be in the y-coordinate.

r(t) = (130t, 4t^2, 6 - 3t - 16t^2)

Since the time of flight is still t = 6/13 sec, we calculate that, in the y-direction, the ball has moved (from t = 0 to t = 6/13):

$y = 4t^2 = 4 (\frac{6}{13})^2 = 0.852 ft$

e) Comparing the first half of the flight (0 ≤ t ≤ 3/13) to the second half of the flight (3/13 ≤ t ≤ 6/13), the movement in the y-direction is:

$4t^2 = 4 (\frac{3}{13})^2 = 0.213 ft\\4t^2_{1} - 4t^2_{2} = 4 (\frac{6}{13})^2 - 4 (\frac{3}{13})^2= 0.639 ft\\$

The ball moves more in the second half of its flight.

f) The batter has to adjust to the ball moving sideways almost eight inches in the last quarter-second of its flight.

g) We will repeat the calculation of r(t), this time with r(0) = (0, −3, 6), v(0) = (130, 0, −3), and a(t) =(0, c, −32). Again, the fact that we are working separately in each coordinate, makes the calculations reasonable.

You can check, that according to the conditions given above

r(t) = (130t, −3 + 0.5ct^2 , 6 - 3t - 16t^2)

At the end of the flight (t = 6/13), the y-coordinate should be 0, so

$-3 + \frac{1}{2}*c*(\frac{6}{13})^2 = 0\\\\c = 28.2 ft/sec^2$

6 0

2 years ago