Integral Help

[MrCodeDude]

Velocity and Acceleration

An airplane taking off from a runway travels 3600 feet 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?

They usually give a starting acceleration speed, so I am able to do these problems. But alas, no such luck 🙁

I know,
S(30) = 3600
S(0) = 0

V(0) = 0
V(30) = ?

A(0) = K
A(30) = K

How do I put that into an equation where I solve for Velocity/Speed?

 

[Kermy]

Been a while since Physics but:

x = x0 + 1/2(v0 + v)t

v0 = intial velocity
x0 = 0

Solve for v

 

[brunswickite]

i think this is correct

x(t) = distance

dx/dt = velocity

dv/dt = acceleration

since acceleration is contant it will be say A*t

the integral of that is 1/2 A*t^2 which would be velocity

the integral of the velocity function would be (A*t^3)/6 = 3600

solve for A, t = 30 seconds

acceleration, A = .8 m/s^2

i am not sure if this is correct.. its been a while since i have done this type of stuff

 

[Oscar1613]

take the indefinite integral of the acceleration eq. to get a general eq for velocity, then take the indefinite integral of that to get the general position eq. then you can use the initial conditions to set up a system of eq's to solve for the unknown coeffs 🙂

 

[MrCodeDude]

Originally posted by: Oscar1613
take the indefinite integral of the acceleration eq. to get a general eq for velocity, then take the indefinite integral of that to get the general position eq. then you can use the initial conditions to set up a system of eq's to solve for the unknown coeffs 🙂

But I don't have a number to start with.

So I'd get:

acceleration (time) = k

integrate that and get:

velocity (time) = .5 k^2 + Vo

position (time) = (1/3) * (k^3) + Vo(t) + So

Position (0) = (1/3) * (k^3) + Vo(0) + 0
Position (0) = (1/3) * (k^3)

How does that help me though?

 

[Oscar1613]

Originally posted by: MrCodeDude

Originally posted by: Oscar1613
take the indefinite integral of the acceleration eq. to get a general eq for velocity, then take the indefinite integral of that to get the general position eq. then you can use the initial conditions to set up a system of eq's to solve for the unknown coeffs 🙂

But I don't have a number to start with.

So I'd get:

acceleration (time) = k

integrate that and get:

velocity (time) = .5 k^2 + Vo

position (time) = (1/3) * (k^3) + Vo(t) + So

Position (0) = (1/3) * (k^3) + Vo(0) + 0
Position (0) = (1/3) * (k^3)

How does that help me though?

you messed up... v(t) = k*t + Vo
p(t) = k/2 * t^2 + Vo*t + Po

you're given Vo and Po = 0, therefore
p(t) = k/2 * t^2
v(t) = k*t
a(t) = k

you're given p(30) = 3600 so just plug that in and solve for k, then solve v(30)

 

[RossGr]

You know

x= .5at^2

and

v=at

so a = v/t

x= .5 (v/t)t^2

x= .5vt
v= 2x/t

 

[Yossarian]

163.6 MPH

 

[brunswickite]

Originally posted by: MrCodeDude

Originally posted by: Oscar1613
take the indefinite integral of the acceleration eq. to get a general eq for velocity, then take the indefinite integral of that to get the general position eq. then you can use the initial conditions to set up a system of eq's to solve for the unknown coeffs 🙂

But I don't have a number to start with.

So I'd get:

acceleration (time) = k

integrate that and get:

velocity (time) = .5 k^2 + Vo

position (time) = (1/3) * (k^3) + Vo(t) + So

Position (0) = (1/3) * (k^3) + Vo(0) + 0
Position (0) = (1/3) * (k^3)

How does that help me though?

i solved it look up

 

[Oscar1613]

Originally posted by: brunswickite

Originally posted by: MrCodeDude

Originally posted by: Oscar1613
take the indefinite integral of the acceleration eq. to get a general eq for velocity, then take the indefinite integral of that to get the general position eq. then you can use the initial conditions to set up a system of eq's to solve for the unknown coeffs 🙂

But I don't have a number to start with.

So I'd get:

acceleration (time) = k

integrate that and get:

velocity (time) = .5 k^2 + Vo

position (time) = (1/3) * (k^3) + Vo(t) + So

Position (0) = (1/3) * (k^3) + Vo(0) + 0
Position (0) = (1/3) * (k^3)

How does that help me though?

i solved it look up

you're a decimal place off 😛