Kinematics: Initial Velocity, Acceleration, Displacement & Time

[SSundee]

Hello All,

A lot of you are probably physics majors so this question will sound dumb to you, but I am just in A.P. Physics 1 in high school... we are learning projectile motion and the teacher is not the best at explaining it. We had our first test a few days ago, everyone bombed it, and he didn't care...:L... anyway, the reason we all tripped up is because we had a lot of conceptual questions instead of plugging in numbers into formulas like we practiced. I know it's really crucial to know these things so here's my question: what are the relationships among initial velocity, acceleration, displacement, and time in 1-D problems? The same goes for 2-D problems (except that there are two numbers for each value in x and y). "Two cars are traveling with the same velocity, they step on the brake at the same time, except one twice as much (with a 2x acceleration). What is the relationship in the time that it takes the two cars to stop?" I guess what I'm going at is what are the correlations between all of these factors, I think it will help me for my tests, and for my understanding of conceptual physics.

Thank you all so much for your help. :)

 

[rtsswmdktbmhw]

It's not a dumb question, I am also not far out of high school and can relate. Hopefully my answer can help a little.

The relationships can be seen by using equations.

For example, using the question you posed: You can use the equation v = u + a*t rearranged to t = (v-u) / a

You know that the two cars have the same initial velocity (u). The final velocity is also the same - because you are measuring the time it takes for cars to stop (i.e. v = 0 m/s). Lastly, you have a(1) = 2*a(2) where a(1) is acceleration of car 1 and is twice that of car 2 (the one and two in brackets are meant to be subscripts).

So the time it takes for car one is t(1) = (v-u)/a(1) = (v-u)/(2*a(2)) and the time it takes for car 2 is t(2) = (v-u)/a(2)

Take the ratio t(2)/t(1) (or the other way round, it doesn't really matter) and you will get t(2)/t(1) = (v-u)/a(2) * 2*a(2)/(v-u)

The (v-u) and a(2) cancels to give you t(2)/t(1) = 2 and this tells you that car 2 (which accelerates half as fast as car 1) takes twice as long to come to rest. If you took t(1)/t(2) you would end up with car 1 taking half as long to come to rest, which is an equivalent answer. Conceptually, it makes sense because if you slow down half as fast you expect to take twice as long (but you'll probably still be expected to show it mathematically!).

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When you are presented with a problem, it is always helpful to first take note of the information that is given. In the example above, you are told initial velocity, final velocity, and ratio of acceleration. (But don't automatically assume that all the information included will be relevant to the solution.)

Then, always ask yourself: what is the question asking you to do? Here the question asks you to find the relationship between time taken by the cars.

Lastly, is there a formula that links the information you are given to an unknown that you are trying to find?

 

[SSundee]

Thank you so much! That helped a lot, I wish my teacher would take time to explain that! Thanks again, I will try to do a bunch of practice to get used to the thought processes. :D