A ball moves in a straight line (the x-axis). The graph in Fig. E2.9 shows this ball's velocity as a function of time. (a) What are the ball's average speed and average velocity during the first 3.0 s?

Question

A ball moves in a straight line (the x-axis). The graph in Fig. E2.9 shows this ball's velocity as a function of time. (a) What are the ball's average speed and average velocity during the first 3.0 s?

Velocity-time graph showing a ball's velocity over 6 seconds, with a constant speed of 4 m/s.

Velocity-time graph depicting a ball's velocity with a step change from 2 m/s to 3 m/s over 3 seconds.

Accepted Answer

Hey everybody today, we're dealing with the problem regarding a velocity versus time graph. So we're being told that a bicycle is coasting down the road in a straight line parallel to the X axis. And the graph below represents the bicycles velocity as a function of time. With this. For being asked to determine both the average speed and average velocity of the bicycle for the 1st 4.5 seconds. So 4.5 seconds lines up right here. So essentially what is the average speed and average velocity of the bicycle during this time frame? That is graft. So to do this, we need to recall that uh average speed, average speed is equal to the total distance, total total distance divided by the change in time again, that is total time. So, in order to get this, well, we know that delta T total is 4.5 seconds. That's what's described in the question as well as in the graph. However, the total distance needs to be calculated using the areas under the under the graph itself. And recall that the area under a velocity and time graph, area under graph is equal to the distance. The change in position. So with this in mind we can say that the distance total right is equal to The area of one rectangle and we'll write this as a rectangle. one which in this case is more of a square And then we'll put this rectangle as area two. Alright, this is Area two. Clean this up a bit. So it'll be the area of one plus the area the next which can be simplified to L one plus two. B one or sorry, Plus my bad. It will be multiplied by W one, wow. Because we're concerned with magnitudes not just of the signs And if we want to write this out we could say this. L one Decances is W one and let us write this as L two And this is with two. Doesn't make sense in just a second because we have to Get the magnitude of L2 into W2 because the area of a square rectangle is just base into height or length into with so expanding that we get, It'll be 2.0 seconds Into four m/s plus 2.5 seconds. Because again, L two goes from 2 to 4.5 seconds. So the length will be 2.5 seconds multiplied by the change in velocity, so it is zero all the way up to seven. So this will be seven m/s. And solving that brings us to eight meters plus 17.5 m half meters, Bringing us to an answer of 25.5 m. So that's the distance traveled during this time period. This time period of 4.5 seconds. So substituting that back in for the average speed. Well, as we mentioned earlier, average speed is just d total over the change in the total change in time. So the sylvie 25. m Divided by 4. seconds. So that'll give us five point seven m per second. Right? So that is our average speed. However, average velocity doesn't require just the magnitude that also includes the areas with their signs. Right? So let's scroll down a bit and write this out. So average velocity and I'll write this green average velocity is equal to the change in total distance or delta X in terms of displacement. Now, because displacement is not the same as total distance. Your displacement depends on your starting and end points and that's where the signs come into play. And negative velocity would mean you're going the opposite direction in this case and it'll be over the total time. That's total time. So again, We know that the time total is 4.5 seconds. So we'll keep that there. No. This would just simply be equal to A one Plus A two. And we actually already have those up here. If we take away the absolute values that give us our magnitude, We get once again eight m for a second. I'm sorry, eight m eight m plus 17.5 m which equals 25.5 m. So substituting that back in 25.5 m gives us an answer of 5.7 m per second. So in this case the average velocity average velocity here as well as the average speed are the same. It is 5.7 m per second and that aligns with answer choice. D. I hope this helps when I look forward to seeing you all in the next one.

A ball moves in a straight line (the x-axis). The graph in Fig. E2.9 shows this ball's velocity as a function of time. (a) What are the ball's average speed and average velocity during the first 3.0 s?

Verified Solution