A particle starts from at and moves with the velocity graph shown in FIGURE EX2.6. (b) What is the object’s position at and 4 s?

Question

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. The given graph shows how a body's velocity changes over time. If the body is starting at X zero equals 7 m at T equals zero seconds. What is its position at 10 seconds? OK. So that's our end goal is we're trying to figure out that if the body starts at X zero. So the initial position is at 7 m when it's at T equals zero seconds, we're asked to figure out what its position is at 10 seconds and that's our final answer and that's our end goal. What we're trying to solve for ultimately. So looking at our graph here, it looks like we start at, at a speed is our Y axis is representing V the speed in meters per second and our X axis is representing the time in seconds. So it looks like our graph starts at negative 3.0 on our V axe and it travels up two, our first peak where it looks like it stops moving for a little bit and it's our going at a constant pace here from 15 to 25 seconds. And then finally, it will decrease down to zero at 30 seconds. However, we're focused at and I'll mark it in red. We're focused at 10 seconds. So what is its position at 10 seconds? So we're trying to figure out where this red dot is? Ok. So we're trying to figure out. So it looks like at 10 seconds, it's going at a velocity of 3.0 m per second. So that's what we're looking at. We're also given some multiple choice answers. Let's read them off to see what our final answer might be. And let's also note that the units are all in meters. So A is seven B is three, C is 10 and D is 15. OK. So to start solve the, to start solving this problem, we need to first off, we need to recall that we can find the position of a particle by examining the area under the velocity time graph. So the area under the velocity time curve between, so we're between zero seconds is less than T, which is less than five seconds and five seconds is less than T and T is less than 10 seconds, we'll cancel out. So the area under the velocity time curve between zero seconds is less than T and T is less than five seconds and five seconds is less than T and T is less than 10 seconds. These will cancel out will cancel each other out because they have equal magnitudes and opposite signs. Therefore, as a result, the position will remain the same as it was initially. So recall that the equation to find the final position is as follow as follows. So XF which represents the final position is equal to X zero, which is the initial position plus the area. So the area under velocity. So under V versus X curve, OK, that was a little cramp, toss a curve. OK. There we go. So let's plug in all of our known variables to solve our final position. So we know that the initial position is 7 m plus one half multiplied by, let's see. Let's put that in front of the, so one half multiplied by five seconds multiplied by negative 3 m per second plus one half. So one half multiplied by, so multiplied. So multiplied by five seconds multiplied by 3 m per second. So as you could see, as we predicted a, a moment ago, XF equals 7 m plus zero because this, I'm gonna bracket it in with a blue border here. So everything in our blue border here will will equal zero when we plug that into a calculator. So therefore, we know without a doubt that XF equals 7 m. So therefore XF equals 7 m at 10 seconds. So that's our final answer. Hooray. We did it and this corresponds with multiple choice answer letter 8 7 m. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

A particle starts from at and moves with the velocity graph shown in FIGURE EX2.6. (b) What is the object’s position at and 4 s?

Verified Solution

Key Concepts

Velocity

Integration

Kinematics