velocity and acceleration

[emylou0109]

Our class is working on review for the AP Test and I can't figure this problem out

An airplane taking off from a runway travels 3600 ft before lifting off. If it starts from rest, moves with constant acceleration, and makes the run in 30 seconds, with what speed does it lift off?

I thought it might have something to do with the integral from 0 to 30 of t dt but I just cant figure it out. The answer is supposed to be 240 feet per second and I can't get the right answer... thanks for the help[/code]

 

[skeeter]

think about the plane's velocity graph ... it's a line starting at (0,0) and ending at (30, v(30)) {you know the graph of velocity is linear because you were given that the acceleration is constant}

the area under this line from 0 to 30 represents the airplane's displacement ... and the area is just a right triangle

area = (1/2)(base)(height)

3600 ft = (1/2)(30 seconds)(v(30))

now, determine the value of v(30).

 

[emylou0109]

Thank you so much for the help. I feel stupid now because that was really easy. Anyway I have one more question and I promise I won’t bother you again this weekend…

A ball is thrown vertically upward from ground level with an initial velocity of 96 ft per second.

a.) How long will it take the ball to rise to its maximum height?
b.) What is the maximum height?
c.) When is the velocity of the ball one-half the initial velocity?
d.) What is the height of the ball when its velocity is one-half the initial velocity?

I think if I could figure out the equation of the velocity graph I could figure out the questions.

 

[soroban]

Hello, emylou0109!

You are probably expected to solve it with Calculus . . .

An airplane taking off from a runway travels 3600 ft before lifting off.
If it starts from rest, moves with constant acceleration, and makes the run in 30 seconds,
with what speed does it lift off?

Let the constant acceleration be \(\displaystyle A\).
Then we have: \(\displaystyle \,a(t) \:=\:A\)

Integrate: \(\displaystyle \:v(t) \:=\:\int A\,dt \:=\:At\,+\,C\)

It starts from rest, so \(\displaystyle v(0) \,=\,0\)
. . We have: \(\displaystyle \:v(0)\:=\:A\cdot0\,+\,C\:=\:0\;\;\Rightarrow\;\;C\,=\,0\)
The velocity function is: \(\displaystyle \,v(t)\:=\:At\)

To find the displacement, integrate: \(\displaystyle \:x(t) \:=\:\int At\,dt\:=\:\frac{1}{2}At^2\,+\,C\)

Assuming its intial position is zero, \(\displaystyle \,x(0)\,=\,0\)
. . we have: \(\displaystyle \:x(0)\:=\:\frac{1}{2}A\cdot0^2\,+\,C\:=\:0\;\;\Rightarrow\;\;C\:=\:0\)
The displacement function is: \(\displaystyle \,x(t)\:=\:\frac{1}{2}At^2\)

It moves 3600 feet in 30 seconds: \(\displaystyle \,x(30)\,=\,3600\)
. . We have: \(\displaystyle \,\frac{1}{2}A(30^2)\:=\:3600\;\;\Rightarrow\;\;A\,=\,8\)

Hence, the velocity function is: \(\displaystyle \,v(t) \:=\:8t\)

At \(\displaystyle t\,=\,30:\;v(30)\:=\:8(30)\:=\:\fbox{240\text{ ft/sec}}\)

 

[soroban]

Hello again, emylou!

A ball is thrown vertically upward from ground level with an initial velocity of 96 ft.sec.

a) How long will it take the ball to rise to its maximum height?
b) What is the maximum height?
c) When is the velocity of the ball one-half the initial velocity?
d) What is the height of the ball when its velocity is one-half the initial velocity?

You are expected to know the free-fall formula: \(\displaystyle \:h(t)\:=\:h_o\,+\,v_ot\,-\,16t^2\)
. . or derive it, know that the acceleration is 32 ft/sec² (downward).

We are given: \(\displaystyle \,h_o\,=\,0,\;v_o\,=\,96\)

The height function is: \(\displaystyle \:h(t)\:=\:96t\,-\,16t^2\)

(a) To find maximum height, differentiate the height function and equate to zero.
. . . \(\displaystyle v(t)\:=\:96\,-\,32t\:=\:0\;\;\Rightarrow\;\;t\,=\,3\) seconds

(b) The maximum height is: \(\displaystyle \:h(3)\:=\:96(3)\,-\,16(3^2)\:=\:144\) feet.

(c) The velocity function is: \(\displaystyle \:v(t)\:=\:h'(t)\:=\:96\,-\,32t\)
. . . When is \(\displaystyle v(t)\,=\,48\)? \(\displaystyle \;96\,-\,32t\:=\:48\;\;\Rightarrow\;\;t\,=\,1\) second

(d) When \(\displaystyle t\,=\,1:\;h(1)\:=\:96(1)\,-\,16(1^2)\:=\:80\) feet