FIGURE EX2.12 shows the velocity-versus-time graph for a particle...

Question

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 (a) What are the particle's position, velocity, and acceleration at t = 1.0s

Accepted Answer

Hey, everyone. Welcome back. In this problem. A sphere undergoes one dimensional motion along the horizontal path. The sphere's velocity as a function of time is shown in the figure below the sphere is located at time T equals zero seconds at Y not equals three centimeters. We're asked to find the position, velocity and acceleration of the sphere at T equals two seconds. Looking at the graph we've been given, we have the velocity in centimeters per second on the Y axis. The time in seconds on the X axis, we have a horizontal curve from zero seconds to three seconds at a value of one centimeter per second and then from three seconds to five seconds, the velocity increases. OK. In a straight line from one centimeter per second to five centimeters per second. We're given four answer choices. Each of them have a different position, velocity and acceleration. The positions are given in centimeters. The velocities are given in centimeters per second and the accelerations are given in centimeters per second squared. Now, this is a velocity time graph. So let's start with finding the velocity. And for that, all we'll need to do is pick out that value off of our graph. So we have the velocity at two seconds is going to be equal to, we go to our graph, we find two seconds. OK? And upwards from two seconds, we have a velocity of one centimeter per second. So the velocity at two seconds is gonna be one cm/s. If we look at our answer choices, We can already eliminate two of the answers. We found the velocity at two seconds to be one cm/s. OK. In B it has a velocity of three centimeters per second. So we know we can eliminate B and D also has a velocity of three centimeters per second at T equals two seconds. So we can eliminate D as well. So we're looking at either answer choice A or B we're left with finding the position and the acceleration. So for the position recall that the displacement or the change in position delta Y is equal to the area under a V T curve. So the change in position from two seconds, two, well, what other point should we choose to look at the change of position? We're given information about the position at T equals zero seconds. OK. We have Y not as equal to three centimeters. So let's look at the change in position from zero seconds to two seconds because we have that information. So the area under the curve is going to be a rectangle. We know how to find the area of a rectangle. The change in position is going to be the position at two seconds minus the position At T equals zero seconds. This is going to equal to the base multiplied by the height of the rectangle. The position at two seconds is what we're looking for. So we have that minus three centimeters, the initial position at T equals zero seconds. And this is equal to the base of our rectangle. Well, it goes from zero seconds to two seconds. So the base is going to be two seconds and the height of our rectangle goes from zero centimeters per second to one centimeter per second. So that's gonna be one centimeter per second. So we have two seconds multiplied by one centimeter per second. On the right hand side of our equation, We have y at two seconds -3 cm is equal to two cm. We can add three cm to both sides to isolate the position Y at two cm and we get a value of five cm. So the position Of our sphere at two seconds is going to be five cm. OK. The velocity is one cm/s. Now, we need to do the acceleration. Now recall that the acceleration is going to be equal to the derivative of the velocity. So A Y is equal to D V Y divided by D T. And when we have A V T graph acceleration is just equal to the slope of the V T curve. The position or the change in position is the area under the curve. The acceleration is the slope of the curve. So our acceleration A Y at two seconds is going to be equal to the slope of this curve at two seconds. Well, this is a horizontal line. We know that a horizontal line has no rise. It doesn't have any change in the Y direction. So this slope is zero. So our acceleration is going to be zero cm/s squared. So we've found our velocity of one centimeter per second. Our position is five centimeters and the acceleration at two seconds is zero centimeters per second squared. This is gonna correspond with answer choice C that's it for this one. Thanks everyone for watching. See you in the next video.

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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 (a) What are the particle's position, velocity, and acceleration at t = 1.0s

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