Postion of an object at a given time?

[badman]

im trying to do this problem involving the postion of a rocket going straight up at a given time.

it says a rocket initially at rest on the ground, accelerates straight upward from rest with constant net acceleration a, until time t(subscript 1) when the fuel is exhausted.

the equation for this is: position at that time=initial position+initial velocity times t(time)+0.5 times 3 times 9.81m/s^2 times T^2

i know initial velocity and position are zero, so all i got left over is the rest of the equation.

a=3g(g being accel. due to gravity) and T(subscript 1)=5.00s
it says to give ur answer in meters. i ended up with 368 meters. but it said that i forgot a numerical factor?

 

[whozum]

x = \frac{1}{2} at^2 = \frac{1}{2} (3*9.8)(5)^2 = 367.5m

 

[badman]

thats what i basically entered but it keeps saying that I am off by a numerical factor? why the hell does it keep saying that.

the height I am supose to give is in this form H (subscript) 3= the answer.

it also says express the answer numerically in meters using g=9.81 m/s^2. just to be clear on that. i don't know what's the problem

 

[dextercioby]

I don't think this has to do with sig figs.

Compute the velocity at the time when all the fuel is burnt.From then on,the rocket goes straight up due to inertia.Gravity will be then the only force acting on it.

Your number ~368 is too small.

Daniel.

 

[whozum]

Are you trying to find the height when the fuel is burnt, or the final height? If its the first, then 368 is the answer.

 

[badman]

it says to look for the maximum height

 

[dextercioby]

Voilà,so i was right.

Daniel.

 

[whozum]

In that case find the velocity at t_1, then use the same equation you gave above with v_0 = v(t_1) and a = -9.8

Solve v_0 - at = 0 for a = 9.8 and v_0 = v(t1) to find where the velocity is 0 (the top). Use that value for t.

This is the distance traveled after the fuel is burned out.

 

[badman]

how would i go about computing the velocity at T_1.?

 

[whozum]

You underwent constnat acceleration of 3*9.8 for 5 seconds.

 

[badman]

thanks for the help. guys.

i have another quetion, but i don't know how to type these leters the right way

The next figure View Figure shows the velocity vectors corresponding to the upward motion of the power ball. Indicate whether its velocity and acceleration, respectively, are positive (upward), negative, or zero.
Use P, N, and Z for positive (upward), negative, and zero, respectively. Separate the letters for velocity and acceleration with a comma.

i tried typing in v p, a n, and other combnations but said it had the wrong number of terms separated by commas. how do i go about typing these terms in?


heres a the figure there referring to.

 

[whozum]

I think its looking for just p, n
Those are correct answers.

 

[badman]

thanks a lot man, that helped with the next problem. i really didnt know how to input these terms in.