Ann and Carol are driving their cars along the same straight road. Carol is located at x = 2.4 mi at t = 0 h and drives at a steady 36 mph. Ann, who is traveling in the same direction, is located at x = 0.0 mi at t = 0.50 h and drives at a steady 50 mph.

c. Draw a position-versus-time graph showing the motion of both Ann and Carol.

Question

Ann and Carol are driving their cars along the same straight road. Carol is located at x = 2.4 mi at t = 0 h and drives at a steady 36 mph. Ann, who is traveling in the same direction, is located at x = 0.0 mi at t = 0.50 h and drives at a steady 50 mph.

c. Draw a position-versus-time graph showing the motion of both Ann and Carol.

c. Draw a position-versus-time graph showing the motion of both Ann and Carol.

Accepted Answer

Welcome back. Everyone in this problem. Sam and Dean are walking in a straight path. Sam is located X equals eight kilometers away from the start point at T equals zero hours. He walks constantly at four kilometers per hour. Dean started from the start point at T equals one hour and walks constantly at six kilometers per hour. Draw a position versus time graph showing the motion of both Sam and Dean. And below, we have a graph showing our time on the horizontal axis and our position in kilometers on the vertical axis. No, if we're going to find our or if we're going to draw sorry, our position versus time graph, let's think about the information we have. Notice that for both some, OK, let's highlight some in blue here and Dean, let's highlight Dean in red. Notice that for both of them, they're walking at a constant rate. Some is walking at a constant rate of four kilometers per hour and Dean is walking at a constant rate of six kilometers per hour. So that means on a position versus time graph, their path or their journey is going to be a straight line. What do we know about straight lines on a, on a graph, we recall that every straight line on a graph has the formula or is written in the form ye equals MX plus B. OK. So if we want to find a position for each of our, for both Sam and Dean, then the position is going to be equal to their initial position X knot plus the gradient of our position versus time graph, which is their speed multiplied by the time, which is the variable on the horizontal axis. If we can find that for both, then we can use the equation to find two points on the graph and thus plot both of their lines. So let's first try to do that with sums position. OK. So let's not do this for a Sam. Now, what do we know about some? Well as our problem told us, some was located eight kilometers away from the start point at T equals zero hours. So some was probably around somewhere here. OK. So for some or rather let's say for Sam's position, OK. We know that. So let me write that here, we know that his initial position is at eight kilometers. OK? And we also know that he's walking at a rate of four kilometers per hour. So his speed V is four kilometers per hour. So if we use that, then that tells us that the formula for Sam's position. OK? Or rather X as X represents wherever his position is so X for sum is going to be equal to eight kilometers is initial position plus four kilometers per hour multiplied by T. In other words, if we want to find some position at any or anywhere, then we can substitute or time t depending on the hours. So we can rewrite this then as eight plus four kilometers for Sam's position. Now, what do we know about Dean? Well, let me talk about Dean's position. OK. Remember that our problem told us if we go back to our problem, we know that Dean started from a start point at T equals one over. So if we were to go to our graph here, T equals one over and walks constantly at six kilometers per hour. So since Dean's start point is at T equals one hour, if we were to go back to our Y intercept here, OK, then the Y intercept for Dean would be negative six. If we were to imagine that, imagine that we extended our line here. OK. So since coin, let's come down back down here, since Dean's Y intercept or the intercept on the position axis is negative six kilometers and he's walking at a rate of six kilometers per hour. OK. Then that means that Dean's position is going to be equal to negative six plus 60 kilometers. Know that we have equations for both Deans and Sam's position, we can plot their graphs. Let's start with some. Now, already, for some, if we're going to plot his graph, we need two points. So for some, let's, let's see if we can find another point for him. And let's try to find a point that preferably has values on the scale at an intersect. So we won't need to estimate how about when the time is three hours? Well, when T equals three, OK, then sums position at three hours is going to be equal to eight plus four multiplied by three. OK. And that's eight plus 12, which is 20 kilometers next. If we were to think about Sam's position at the eight kilometers. Ok? Or at eight hours, sorry, when T equals eight hours, then his position is going to be equal to eight plus four multiplied by eight. That's eight plus 32 which is 40 kilometers. So since we have 340 ok? And sorry, 320 my apologies. 3 28 40 then we can go ahead and plot for Sam's position, ok. So at three hours, Sam is going to be 20 kilometers away while at uh eight or some is gonna be 40 kilometers away. Since we have two points, now we can go ahead and plot our line. So if we go ahead and do that here by connecting our line, OK. Oops, I'm going to try and draw this properly. OK. Then notice that this shows this is the graph that shows Sam's motion. OK? So we have our motion for s next. Let's try to find our position for Dean. OK. So we already have one point. OK. At 10, let's say we wanted to find his position at six hours. OK? So when T equals six for Dean and the time is six hours, then Dean's position is going to be equal to negative six. OK. Plus six, multiplied by six, that's negative six plus 36 which equals 30 kilometers. So when t equals six hours, the position is 30. So that gives us the 0.6 30. So when we go back to our graph, we can go ahead and plot 630 which would be right here. And now we have both of those, we can go ahead and plot our graph by connecting those two points. OK? And now when we do that, notice that our graph then is gonna look something like that. OK? And it's a good thing, we use both of those points because my negative six was an estimate based on the graph. It wasn't an exact value. But now we have those that coi coincided with the units of the graph. So this grid line represents Dean's position or Dean's motion over time. Therefore, this would be the position versus time graph showing the motion of both Sam and Dean. Thanks for watching everyone. I hope this video helped.

Verified Solution

Key Concepts

Position vs. Time Graph

Constant Velocity

Relative Positioning