A box moving with an initial speed v is accelerated horizontally. If x is measured in [m], v in [m/s], a in [m/s 2 ], t in [s] which of the following equations is correct for solving the distance x ?

Hey, guys. Let's check out this practice problem. Hopefully, you got a chance to try it on your own. We have a box moving with an initial speed v, and we're told the units for for x, v, a, and t, and we're gonna figure out which one of these equations is correct for solving for the distance. So really, we're just gonna figure out which one of these equations is dimensionally consistent. So let's take a look at a. So for a, I'm just basically gonna replace the variables with their units. I know that x is given to me in meters or that's the unit for that. I know that v is meters per second. A is meters per second squared. So I have a over t squared, so I'm gonna replace those units. I've got meters per squared on, meters per second squared on the top and in the bottom I've got another factor of seconds squared on the bottom. So if you simplify this and start canceling out your units or multiplying and dividing, what you're gonna get is you're gonna get meters on the left. And then if you have meters per second squared on the bottom and you divide by another second squared on the denominator, what you're gonna get is you're gonna get meters per seconds to the 4th power. So this is basically gonna be on the left and right side. And now we just check if they're equal to each other. Obviously, they're not. So that means that this equation cannot be cannot possibly be correct. It's dimensionally inconsistent. So let's just quickly go through b and c again. So let's just go through, B. This is gonna be meters on the left. And then my velocity. So velocity is gonna be in meters per second. So I have meters per second plus And now I have 1 half we don't really care about those factors or minus signs and then the acceleration is gonna be meters per second squared and then I have time, which is just seconds. So now we just have to figure out if this is dimensionally consistent. Well, I've got meters per second over here, and I'm gonna add it to meters per second squared times seconds. So one of these factors will cancel out. So what I end up getting when I simplify this is I got meters in the left and I get meters per second plus meters per second. So, basically, if you're adding 2 of the same unit together, then this will simplify to meters on the left and then a meter per second plus another meters per second is just a meters per second. You can add together 2 of the same unit and that's perfectly okay and it just condenses down to 1 unit. So but, but the fact remains that the still is not actually equivalent on both the left and the right side. There is an inconsistency here. So we don't have the same units on the left and right side. So therefore, this equation is outs. So a and b are out. So that means this must be answer choice c. But just to be sure, let's just go ahead and do it really quickly. If you do the same thing for c, we're gonna have meters on the left side, and this is gonna be v times t. So my velocity is meters per second times seconds seconds plus, and then, again, we don't care about the one half, it's just factor, meters per second squared times second squared. So what's gonna happen is, what we're gonna see is that seconds will cancel out with seconds in the first term, and then second squared will cancel out with second squared in the second term. So basically what we're gonna get, because we're gonna get meters on the left side equals meters plus meters. And so, we can simplify this and say meters is again, if you add 2 2 together of the same kind, then it just simplifies to 1, and we're gonna get meters equals meters on the left and right side. This is dimensionally consistent and therefore, it must be your answer. Alright guys. That's it for this one. Let me know if you have any questions.

Physics with Algebra 때문에 힘들어하고 계신가요?

A box moving with an initial speed v is accelerated horizontally. If x is measured in [m], v in [m/s], a in [m/s2], t in [s] which of the following equations is correct for solving the distance x?

다음 시청

A box moving with an initial speed v is accelerated horizontally. If x is measured in [m], v in [m/s], a in [m/s²], t in [s] which of the following equations is correct for solving the distance x?