# Acceleration formula

1. Oct 9, 2006

### bayer

Hello!. I am in trouble with my homework.
I got next data:
The accelration is 90m/s2(that 2 means squared).
Then the distance is 2km(2000m).
I have to find how high speed will he reach in 2km-s(thats the distance i wrote up there).
The task text is like this:

Rocket starts with accelration 90m/s2(that 2 means squared). How high speed he reaches after 2km-s. What kind of formulas i need to solve this little problem.

Thank you.

2. Oct 9, 2006

### bayer

Also i will add that 90m/s2 is probably g(in some cases a), the 2km is S(i think) And i need to find t(time) and Vo(the start speed) to find the V(the speed at end) . Or anyone has any other ideas

3. Oct 9, 2006

### Office_Shredder

Staff Emeritus
The 90 m/s2 is probably the upwards acceleration of the rocket, seeing how g is only 9.81

4. Oct 9, 2006

### bayer

Oh thank you =)

5. Oct 9, 2006

### drpizza

Look at a picture of a rocket on the launch pad and think about the initial velocity.

You don't need to find the time. While I don't advocate "plug and chug", there is a formula that relates the initial velocity, final velocity, acceleration, and displacement.
The formula you're looking for is located in Doc Al's second post:

Last edited by a moderator: Apr 22, 2017
6. Oct 9, 2006

### bayer

ok i did like this:
1st t=Vo / g(a)= 90 : 9.8 m/s2=9.184s
V=s/t
V=2000m : 9.18m = 217.86 m/s

or

v=Vo +at
v=90 m/s2 + 9.8m/s2*9,184s=180 m/s

which one is the right one?

7. Oct 9, 2006

### Vesper89

You don't need the time at all....

You do know that V1= 0 m/s

(use V2²= V1² + 2aΔd)

8. Oct 9, 2006

### bayer

Yes, but we have nt learned this kind of formula yet, so i need to choose one of the previous ones, but wich one?

9. Oct 9, 2006

### drpizza

Well, first lets look at your answers and see if they're reasonable. You have an acceleration of about 90 m/s^2. The final velocity of your first answer was 217 m/s, and the final velocity via your second method was 180 m/s.
For the second case, it would take 2 seconds since the acceleration means the velocity is increasing at a rate of 90 m/s per second. And, if the final velocity was 180 m/s, then the average velocity worked out to be (0m/s + 180m/s)/2 = 90m/s. And in 2 seconds, with an average velocity of 90 m/s, you would travel 180 meters. Unfortunately, that's quite a bit off from 2000 meters. For the first case, it would only take a little more than 2 seconds (less than 2 1/2 seconds). Following similar estimations, you're still a far way from 2000 meters.

It's kind of odd that you haven't seen final velocity squared = initial velocity squared + 2 times acceleration times change in displacement.

Without that formula, it's going to take a lot of work with 3 different equations, leaving variables in as you go.

10. Oct 9, 2006

### bayer

(use V2²= V1² + 2aΔd)

But how i find v2 and v1, whats that 2aΔd. We really did nt learn it

11. Oct 9, 2006

### bayer

And also i noticed that took 9.184 second and the distance was 2000m and the speed was 180 m/s. 180*9,184=1653m, but it accelrates 90 m/s. so it takes about 2 seconds. 2+9,184=11,184
11,184*180=2013.12 m. So it should be logical.

12. Oct 9, 2006

What did you mean by '2 km-s'? Did you mean the distance of 2 km? If so, use the formula for distance with constant acceleration first: $$d = d_{0}+v_{0}t+\frac{1}{2}at^2$$. The initial distance and velocity equal zero, so, the formula turns into $$d=\frac{1}{2}at^2$$. Now, you know the distance d, and the acceleration a, right? So, plug in the values, and solve for t to obtain the time the rocket reaches 2000m. Further on, use that value t in the next formula: $$v = v_{0} + at$$. Again, since the initial velocity equals zero, only use v = at to obtain the velocity the rocket has at 2000m.

13. Oct 9, 2006

### drpizza

edit: I'm slapping my forehead at the moment (had one of those duh moments) As radou showed above, you can find the time first. For a moment, I had forgotten that you probably already had that equation. I typically derive that equation in class after I derive the equation below:
--------------
Here is the derivation of that formula from the basic formulas that you have.

$$v_f=v_0+at$$ subtract $$v_0$$ from both sides

$$v_f-v_i=at$$

Now, you know that $$t=\frac{d}{v_{avg}}$$ substitute this for t in the equation above.

$$v_f-v_i=\frac{ad}{v_{avg}}$$

And, you know that $$v_{avg}=\frac{v_0+v_f}{2}$$

substitute for $$v_{avg}$$ in the equation above.

$$v_f-v_0=\frac{ad}{\frac{v_0+v_f}{2}}$$

eliminating the fraction in the fraction by multiplying the numerator and denominator by 2, gives:

$$v_f-v_0=\frac{2ad}{v_0+v_f}$$

multiply both sides by $$v_0+v_f$$ and you get (after FOIL)
$$v_f^2-v_0^2=2ad$$

Rearrange this as $$v_f^2=v_0^2+2ad$$

and, finally, take the square root to find your final velocity. If you follow this, you're going to have unknown variables each step of the way which you will have to keep substituting in. Definitely a pain in the neck to do, but I can't think of another way (perhaps there's a quicker derivation though) besides this to do your problem. I suspect that if your teacher hasn't covered this 5th or 6th formula yet, then perhaps it was a mistake that he gave you this problem already. (I do so every year intentionally, just so I have a reason to go through this derivation; but never as a homework problem.)

Last edited: Oct 9, 2006
14. Oct 9, 2006

### bayer

Thank you for your help, i try to figure something out.

15. Oct 9, 2006

### drpizza

just follow the two steps radou gave you. From one of your posts above, I think you're having a problem with different letters being used.

For displacement, sometimes d is used, sometimes s, and sometimes x.
Likewise, for the initial velocity, sometimes it's $$v_0$$ and in other books, it's $$v_i$$