FIGURE EX2.9 shows the velocity graph of a particle. Draw the particle's acceleration graph for the interval .

Video transcript

Hey, everyone in this problem, the graph below represents the performance of an amateur sprinter during a short race. We're asked to make a graph showing the sprinter's acceleration over time. The graph were given shows the velocity V as a function of time. OK. So we have the velocity in meters per second on the Y axis. The time in seconds on the X axis, we're given four answer choices A through D, they each show a graph of the acceleration In three segments and they all show different accelerations for different times. So let's work through this. Let's go back up to our velocity time graph and I wanna find the acceleration. So let's think about the acceleration and it's a relationship to the velocity or call it the acceleration A X is equal to the derivative of the velocity with respect to time. So we have A X is equal to D V X divided by D T. And when we have a velocity time graph, this is just going to be equal to the slope of the V T curve, the velocity time curve. So to find the acceleration, we just want to find the slope of this graph. Now, our graph has three segments in it From zero seconds up to six seconds. We have this velocity in this positive straight upward line From six seconds to eight seconds. We have this decreasing straight line And then from eight seconds to 10 seconds, we have a horizontal line. So the acceleration is going to be different for each of those segments, but it's going to be consistent throughout those segments. So let's start by finding the acceleration for each of those segments. And then we can go ahead and draw a graph. So for the first segment which runs From T equals zero seconds, The T equals six seconds, we wanna find the acceleration A X. We said that this is equal to the slope of the curve. Recall when you're looking for slope, you can look at the rise over the run. In this case, we have V X on our Y axis. So the rise is going to be the change in the V X and the run is going to be the change in time, which is on our X X. So we have delta V X divided by delta T. We go to our graph. We see that at zero seconds, the velocity is zero m/s And at six seconds, the velocity was nine m/s. So we can use those values in our equation to find the acceleration for this period. We have the change in velocity is the final velocity nine m/s minus the initial velocity, zero m per second divided by the change in time, which is the final time of six seconds, the minus the initial time of zero seconds. So we end up with nine m per second divided by six seconds, which gives an acceleration of 1.5 m per second squared. And because the velocity is a straight line, the slope is the same for that entire segment. And so from 0 to 6 seconds, our acceleration is going to be that same 1.5 m per second squared, moving to the next segment Which runs from six seconds, 28 seconds. Again, the acceleration is going to be delta V XX divided by delta T the change in the Y axis divided by the change in the X axis, the initial time Or sorry, the initial velocity when we're at six seconds. OK. We can look at our graph and see that that is going to be 9m/s. And according to our graph at eight seconds, the velocity is going to be eight m/s. So working out our acceleration, the final velocity Is eight m/s. The initial velocity was nine m per second. So we have eight m per second minus nine m per second divided by the change in time, eight seconds minus six seconds. And we end up with negative one half meters per second squared. That makes sense. That line is sloping down to the right. So we have a negative slope, a negative acceleration. And finally from eight seconds, 2, 10 seconds, we have that our acceleration again will be the change delta V X divided by delta T change in velocity or the change in time. According to our graph, this is a horizontal line in this segment. So the initial velocity and final velocity are the same. They are both eight m per second. So we get eight m per second minus eight m per second, divided by 10 seconds minus eight seconds in the numerator, eight m per second minus eight m per second gives us zero. And so we get zero m per second squared, which makes sense because this was a horizontal line on our graph, we know that a horizontal line has a slope of zero. So we have these three accelerations for three different periods. Now, we can draw our graph, we're gonna draw our axis here and we need to separate the T axis. OK? Which is the X axis into 10. So 246, 8, 10. Now the Y axis is gonna represent our acceleration A X in meters per second squared. And we need this to go from negative and a half all the way up To positive 1.5. And now let's plot hour Values. So from zero seconds all the way up to T equals six seconds, we have an acceleration of 1.5 m per second. So we are gonna draw a horizontal line between zero seconds and six seconds. That is at 1.5 m per second squared From six seconds to eight seconds. We have an acceleration of negative one half meters per second squared. So again, we're gonna draw a horizontal line between six seconds and eight seconds. It has a y value of -0.5. And then from eight seconds to 10 seconds, we have an acceleration of zero m per second. And so we're gonna have a horizontal line from 8 to 10 seconds that lies along the X axis. And that is the graph that shows the acceleration in this situation. If we compare this with the answer choices Again, we had a horizontal segment from 0-6 seconds with an acceleration of 1.5 m per second squared. Then another horizontal segment from six seconds to eight seconds at negative 0.5 m per second squared and a final horizontal segment from eight seconds to 10 seconds along the X axis at A X equals zero. So this corresponds with answer choice D this matches the graph that we drew. Thanks everyone for watching. I hope this video helped see you in the next one.