Simple projectile motion question

[ashclouded]

Homework Statement

A missle is fired vertically up from a ground with an initial velocity of 20m/s. The acceleration at any time is given by dv/dt = -10 + t/2 m/s^2.
Find the height of the missile after two seconds
Find the height after four seconds
the direction the missile is traveling when t = 4

Homework Equations

Displacement = x
Velocity = dx/dt
acceleration = dv/dt

The Attempt at a Solution

dv/dt = -10 + t/2 m/s^2
dx/dt = -10t + t^2/4 + c

found c to be 20

x = -5t + x^3/12 + 20t + c

when t = 0 displacement = 0
c = 0

-5t + x^3/12 + 20t

then I don't know what to do

 

[SteamKing]

Homework Statement

A missle is fired vertically up from a ground with an initial velocity of 20m/s. The acceleration at any time is given by dv/dt = -10 + t/2 m/s^2.
Find the height of the missile after two seconds
Find the height after four seconds
the direction the missile is traveling when t = 4

Homework Equations

Displacement = x
Velocity = dx/dt
acceleration = dv/dt

The Attempt at a Solution

dv/dt = -10 + t/2 m/s^2
dx/dt = -10t + t^2/4 + c

found c to be 20

x = -5t + x^3/12 + 20t + c

when t = 0 displacement = 0
c = 0

-5t + x^3/12 + 20t

then I don't know what to do

You've managed to mangle your calculus here, which is why you get confused.

According to the OP, dv/dt = -10 + t/2 m/s²

If you separate the variables and integrate, you get

∫ dv = ∫ (t/2 - 10) dt, or

v = t²/4 - 10t + C m/s

According to the OP, v @ t = 0 is equal to 20 m/s, so C = 20

But, v = dx/dt = t²/4 - 10t + 20, so separating variables and integrating again we get:

∫ dx = ∫ [ t²/4 - 10t + 20] dt which gives

x = ∫ [ t²/4 - 10t + 20] dt

which I'll leave you to work out. You should be able to answer the questions about the height of the rocket very easily.

 

[ashclouded]

Sorry about the multiple post, thanks for the help but