Trouble Deriving a Constant Acceleration Formula

[berenmacbowma]

Q: How long does it take a car to cross a 30 m-wide intersection after the light turns green, if it accelerates from rest at a constant 2.00 m/s^2?

Attempt: V=V0 + at; x-x0/t=v0+at; x-x0=v0+at^2; x=v0+x0+at^2; 30=0+0+2t^2; 15=t^2.

The square root of 15 ended up being my final answer, when the real answer was closer to 5.48 seconds. I just started physics, so I need to know what I personally did wrong when USING the formula I attempted to use. Giving me a different formula won't be helpful, because it won't tell me what I did wrong, and because my book has a whole chart of formulas I could refer to. If someone could explain to me how I derived that formula incorrectly, it would be a load of help. Thank you!

 

[PhanthomJay]

You derived the formula incorrectly by assuming that velocity at time t is displacement/time. That it is not true when velocity is not constant. To compound the problem, your algebra is wrong in equation 3, but it doesn't matter since equation 2 is wrong.The kinematic equations are best derived form the calculus, check out a good website for this. You might want to refer to the chart of formulas in the book, and you will discover that none of them are the same as the one you derived.

 

[Muphrid]

Q: How long does it take a car to cross a 30 m-wide intersection after the light turns green, if it accelerates from rest at a constant 2.00 m/s^2?

Attempt: V=V0 + at; x-x0/t=v0+at

Right there; that does not follow. The only way you're going to be able to make the correct step is through integration, which if you're not familiar with isn't going to be possible at all.

 

[rcgldr]

You can avoid integration by noting that in a constant acceleration situation, average velocity for any time period equals half the sum of the starting and ending velocities.

average veloctiy = av = 1/2 (v0 + v1)
v1 = v0 + at
av = 1/2 (v0 + (v0 + at))
av = 1/2 (2v0 + at)
av = v0 + 1/2 a t

then distance equals x0 + average velocity x time:

x = x0 + av t = x0 + (v0 + 1/2 a t) t
x = x0 + v0 t + 1/2 a t²

 

[PJC01]

There are a few things that could have gone wrong in your derivation of the constant acceleration formula. One possibility is that you may have made a mistake in your algebraic manipulation. It can be helpful to double check your work and make sure you are following the correct steps to solve for time.

Another possibility is that you may have used the wrong initial velocity (V0) in your equation. Since the car is starting from rest, the initial velocity should be 0 m/s. Using the wrong initial velocity can lead to a different answer.

Additionally, it is important to pay attention to the units in the equation. In this case, the acceleration is given in meters per second squared (m/s^2), so the time should be in seconds (s). If you use the square root of 15 as your answer, you are using the units of meters instead of seconds, which would result in an incorrect answer.

Lastly, it is always a good idea to check your answer using common sense. In this case, if the car is accelerating at a rate of 2.00 m/s^2, it should take less than 15 seconds to cross a 30 m intersection. Therefore, your answer of 5.48 seconds is more reasonable.

In general, when working with formulas, it is important to carefully follow the steps and pay attention to units to ensure an accurate result. It is also helpful to double check your answer using common sense or by plugging it back into the original problem to see if it makes sense. I hope this helps you in your understanding of the constant acceleration formula.