Change the chapter
Question
A powerful motorcycle can accelerate from rest to 26.8 m/ s (100 km/h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time?
1. $6.87\textrm{ m/s}^2$
2. $52.3 \textrm{ m}$
Solution Video

OpenStax College Physics Solution, Chapter 2, Problem 28 (Problems & Exercises) (2:33)

Rating

3 votes with an average rating of 4.

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

Video Transcript
This is College Physics Answers with Shaun Dychko. This motorcycle starts from rest— we are told— so that means the initial velocity is zero; it has a final velocity of 26.8 meters per second and it accelerates during a time of 3.90 seconds. So we write down all this information that we are given. Part (a) asks us to find the average acceleration so we take the change in velocity, divide that by time, that's equation 10 in chapter 2. So that's 26.8 meters per second— final velocity— minus the initial velocity of zero divided by 3.90 giving an acceleration of 6.87 meters per second squared. And part (b) asks us how far did it go and there are two ways that we could answer this question; you might consider equation [2.53] which says that the final position is the initial position which we assume is zero, we could have written that up here as well, plus the initial velocity times time plus one-half times acceleration times time squared and this term would be zero, this term would be zero— this term is zero because it has an initial velocity of zero— and we found the acceleration in part (a) and we could plug that into here multiply by time squared, which we know, and get our answer. But I discourage using this formula because you are using a number that you calculated in this formula and if you made a mistake with this calculation, subsequent calculation's using that will also be a mistake. So let's use a formula that uses the raw data that we are given instead. So I'm using equation [2.50] because we can figure out the average velocity because we know what the initial and final velocities are and we are given the time and we know that the initial position is zero so this uses raw data instead of using numbers that we have calculated before. So we plug in this formula for average velocity here and so we have that the final position is the initial position plus the average velocity times time. So that's 0 meters—initial position—plus 0 meters per second—initial velocity—plus 26.8 meters per second—final velocity— divided by 2 times 3.90 seconds giving 52.3 meters will be the distance that it travels.