FIGURE EX2.12 shows the velocity-versus-time graph for a particle moving along the x-axis. Its initial position is at x0 = 2m at t0 = 0s (b) What are the particle's position, velocity, and acceleration at t = 3.0s

Question

Velocity-time graph showing a particle's motion along the x-axis over time.

Accepted Answer

Hey, everyone in this problem, the velocity versus time data of a remote control car collected during a high school experiment is shown in the figure below the car moves along a straight line initially at time T equals zero seconds. The car was located at X knot equals one m. We're asked to find the position, velocity and acceleration of the car at T equals 40 seconds. Now, the graph we're given, we have velocity uh the horizontal velocity given in meters per second along the Y axis or the vertical axis. And along the horizontal axis, we have the time in seconds At zero seconds. We have a velocity of 3m/s and that remains constant until seconds. And then at 20 seconds, the velocity increases linearly until about 4.4 or 4.5 m/s. At time, 50 seconds, we're given four answer choices. Option A, the position X at 40 seconds is equal to 100 and 20 m. The velocity at 40 seconds is equal to three m per second. The acceleration is equal to zero m per second squared. Option B The position is equal to 121 m. The velocity is equal to four m per second and the acceleration is equal to 0.5 m per second squared. Option C The position is 130 m. The velocity is three m per second and the acceleration is zero m per second squared. And option d the position is 131 m. The velocity is four m per second and the acceleration is 0.5 m per second squared. Now we're given a velocity time curve. So let's start with finding the velocity. OK. This is exactly what the graph shows. So we're just gonna need to pick that particular velocity out at that time point. And so the velocity V X at 40 seconds, it's going to be equal to And we find 40 seconds along the horizontal axis, we go straight up until we hit our velocity curve And this corresponds to a velocity of four m/s. So the velocity at 40 seconds is going to be equal to four m/s. Mhm. Now, we can already eliminate two answer choices. Option A and option C have velocities of three m per second. We know that those are incorrect. So we know that we're looking at answer choice, either B or D. OK. So we've done the velocity, let's move to the position. We have to figure out how is the position related to a velocity time curve will recall that the change in position, delta X is going to be equal to the area under a V T K. Now, we wanna find the area Under this curve all the way up to the time point we're interested in. Now, the time point we're interested in is 40 seconds. So we wanna find the area between our curve to 40 seconds and we can break this up into two shapes. Now, we have a triangle, then we have a large rectangle. So the change in position Between 0.40 seconds is going to be the length times the width of that rectangle Plus one half, multiplied by the base, multiplied by the height of that little triangle. Yeah, we're just finding the area of those shapes under that curve. Now our change in position. Well, we can write this as the final position X F hm, Which is gonna be the position at 40 seconds. So let's write this as X At 40 seconds - Our initial position. X. Well, we know X not, We're gonna be able to calculate this the area of these shapes. And so we can solve for this position at 40 seconds that we're looking for. Now the length of this triangle. Yeah, we're going from zero seconds all the way up to 40 seconds. That's gonna give us 40 seconds multiplied by the width. We go from zero m per second up to three m per second. So we're multiplying by three m per second. Then we're gonna add the area of the triangle. We have one half multiplied by the base of the triangle. The base goes from 20 seconds to 40 seconds. So that base is 20 seconds long, multiplied by the height we go from three m per second to four m per second. So we have a height of one m per second. Now, we have X at 40 seconds a position at 40 seconds minus the initial position, which were given in the problem as one m Is equal to 40 seconds, multiplied by 3m/s, gives us 120 m plus one half, multiplied by 20 seconds, multiplied by one m per second gives us 10 m. Now, we can move this one m to the right hand side by adding it. So we get 100 and 20 m plus 10 m plus one m, Which gives us our position at 40 seconds equal to 131 m. So we found our velocity to be four m/s. And now we've found that the position at 40 seconds is equal to m. All right. So if we look at our answer choices, we can narrow it down and say that the correct answer here is D because the position and the velocity match the answers we found and they're the, that's the only choice that does match both of those. But let's go ahead and calculate the acceleration as well. We wanna make sure that we didn't make any mistakes. Um And it's great practice to see how to solve this problem. So we want to calculate the acceleration and how is the acceleration related to the position or sorry? The velocity time curve? Well, we call it the acceleration is the derivative. OK. So D V X ID T the derivative of the velocity which is represented in our graph as the slope. So our acceleration is gonna be the slope of the V T curve at 40 seconds. Yeah. So our acceleration at 40 seconds, let's take a look at our diagram. Now at 40 seconds, we're along this straight inclined. OK? We have a positive slope section of our curve the slope of this line because it's a straight line is gonna be the same anywhere along that line. So we go, we can go ahead and calculate slope using our rise over, right. OK. So the acceleration is gonna be equal to the rise, which is the change in velocity over the run, which is a change in time. OK. So the change in the vertical axis divided by the change in the horizontal axis, we wanna choose points that are easy to see clearly what values are at. So at 40 seconds, we have a velocity of four m/s And at 20 seconds, we have a velocity of 3m/s. So let's use those two time points that are along that same curve. OK? So the change in velocity is gonna be that final velocity 4m/s Minour the initial velocity, three m/s Divided by the change in time, OK? was the velocity at 40 seconds -3m/s was the velocity at 20 seconds. So we subtract 20 seconds. So we have four m per second minus three m per second, divided by 40 seconds minus 20 seconds. We get one m per second divided by seconds, Which gives an acceleration of 0.05 m/s squared. So we found that the velocity Of the car is four m/s at 40 seconds. The position is 131 m and the acceleration is 0.5 m per second squared. And this does indeed correspond with answer choice. D Thanks everyone for watching. I hope this video helped see you in the next one.

FIGURE EX2.12 shows the velocity-versus-time graph for a particle moving along the x-axis. Its initial position is at x0 = 2m at t0 = 0s (b) What are the particle's position, velocity, and acceleration at t = 3.0s

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Key Concepts

Velocity

Acceleration

Position Function