Time taken to travel x meters with constant acceleration

AI Thread Summary
A user is trying to calculate the time it takes for a rocket, accelerating at 20 m/s², to reach a height of 1379.8 meters. They initially confuse acceleration with velocity and seek clarification on the correct equations to use. After confirming the initial velocity is zero, they apply the kinematic equation for displacement but mistakenly introduce a negative sign during calculations, leading to a square root of a negative number. Other users point out the error in sign and confirm that the approach to solving the problem is fundamentally sound. The user ultimately resolves the confusion and arrives at a reasonable time estimate of approximately 11.746 seconds.
the_cloud
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Hi, I'm trying to solve something that I'm very sure is simple, I just can't seem to find the right equation.

A rocket is traveling straight up, with an acceleration of 20m/s^2. I need to know how long it takes to reach a height of 1379.8m (this is the second part of a question).

I am confused about the acceleration. Does the fact that it's traveling at 20m/s mean that after 2 seconds it's traveling at 400 m/s and at 3 seconds it's traveling at 800 m/s?

And to work out the time it takes to travel that far the equation would be t = 1379.8 / ?

Thanks
 
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It's not "travelling at 20 m/s". The acceleration of 20 m/s^2 means that's the velocity increases by an additional 20 m/s every second. But you need another "given" number to solve this problem, so was it also given that it began at rest, so that the initial velocity can be set to zero? If that's number is also given, you have enough information. Then you need the topic called "kinematics", which provides four formulas. One of those formulas mentions these variables: displacement, initial velocity, acceleration and time.
 
a=\frac{dv}{dt}
dv={a}*{dt}
v=at+v_0
v=\frac{dx}{dt}
dx={v}*{dt}
x=x_0+{v_o}{t}+\frac{1}{2}{a}{t}^2
 
Yep, the starting velocity was 0. I'll look into kinematics and see if I can make sense of those formulas.
 
OK, so I've had a go. Can someone please check my answer and see if it's correct? (Sorry for the crappy formatting, I'm not sure how to do sub\superscripts)

initial velocity = 0;
acceleration = 20m/s^2;
displacement = 1379.8;

d = vi * t + 0.5 * a * t^2
1379.8 = 0m/s * t + 0.5 * 20m/s^2 * t^2
1379.8 = (0m) * t + 10m/s^2 * t^2
1379.8 / -10 = t^2
SQRT(-137.98) = t //?
t = 11.746 seconds

I'm not really sure how I ended up with a negative in my square root, but using the positive I get a number that seems reasonable.
 
Your answer is fine.
the_cloud said:
I'm not really sure how I ended up with a negative in my square root,
When you divided both sides by 10, for some reason you added a negative sign. (Perhaps you were mixing that up with subtracting 10 from both sides?)
 
Ah, yep I did. Thanks for everyones help.
 
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