Change the chapter
Question
Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at $t = 30.0 \textrm{ s}$ is 0.24 m/s.
Question Image
<b>Figure 2.62:</b> Positon vs time graph.
Figure 2.62: Positon vs time graph.
Question by OpenStax is licensed under CC BY 4.0.

$v = 0.217 \textrm{ m/s}$

Solution Video

OpenStax College Physics Solution, Chapter 2, Problem 61 (Problems & Exercises) (1:18)

Sign up to view this solution video!

Rating

No votes have been submitted yet.

Quiz Mode

Why is this button here? Quiz Mode is a chance to try solving the problem first on your own before viewing the solution. One of the following will probably happen:

  1. You get the answer. Congratulations! It feels good! There might still be more to learn, and you might enjoy comparing your problem solving approach to the best practices demonstrated in the solution video.
  2. You don't get the answer. This is OK! In fact it's awesome, despite the difficult feelings you might have about it. When you don't get the answer, your mind is ready for learning. Think about how much you really want the solution! Your mind will gobble it up when it sees it. Attempting the problem is like trying to assemble the pieces of a puzzle. If you don't get the answer, the gaps in the puzzle are questions that are ready and searching to be filled. This is an active process, where your mind is turned on - learning will happen!
If you wish to show the answer immediately without having to click "Reveal Answer", you may . Quiz Mode is disabled by default, but you can check the Enable Quiz Mode checkbox when editing your profile to re-enable it any time you want. College Physics Answers cares a lot about academic integrity. Quiz Mode is encouragement to use the solutions in a way that is most beneficial for your learning.

Calculator Screenshots

OpenStax College Physics, Chapter 2, Problem 61 (PE) calculator screenshot 1
Video Transcript
This is College Physics Answers with Shaun Dychko. To find the instantaneous velocity at 30 seconds, we need to take a tangent line to the curve at this point. So the tangent line will touch the curve only at this point and not at any other point, although it's hard to really see that because this red line is nearly a straight line. But I've drawn a blue line here as my tangent line. So then we find the slope of the tangent by finding out how much it rises, and that's a change in position of about 17 meters at the top minus 4 meters at the bottom. Then we're going to divide that rise by the time interval over which this change of position occurs. So it starts at about 10 seconds up to about 70 seconds and so that's a time interval of 10 minus 70, 70 minus 10 which is 60 seconds. So we have the velocity then is 13 meters divided by 60 seconds which is 0.217 meters per second which is close enough to 0.238 meters per second that we were meant to verify. We don't expect to get exactly that number because we're doing an approximation here visually using a graph. So, close enough though.

Comments

Submitted by xcoonrod99 on Sun, 06/09/2019 - 11:16

How do we get the tangent line?