Question

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of $0.500 \textrm{ m/s}^2$ for 7.00 s. (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

- $15.0 \textrm{ m/s}$
- $5.27 \textrm{ s}$
- $\Delta t = 4.2 \textrm{ s}$

$\Delta x = 49 \textrm{ m}$

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Video Transcript

This is College Physics Answers with Shaun Dychko. A cyclist has an initial speed of 11.5 meters per second and they accelerate for 7 seconds with an acceleration of 0.500 meters per second squared and the question is in part (a), what is their final speed after this period of acceleration? So we have 11.5 meters per second plus 0.50 meters per second squared multiplied by 7 seconds giving us 15.0 meters per second and this is equation [2.52] here by the way. Now in part (b), the question gets more interesting; it says that the cyclist accelerates for 7 seconds and then continues for another period of time with zero acceleration. So at a final speed of 15, they go from this position,

*x 1*, that we don't know to the finish line, which is at*x 2*, 300 meters away from the beginning. And so we are gonna compare the total time in two different scenarios; we have this total time*t 2*when the cyclist reaches the finish line after having done this period of acceleration and we are gonna compare that with*t*which I give a subscript 'const' constant which is down here and when you take the difference between the two, the constant time versus the time with acceleration, which I call*t 2*, and we get that difference 5.27 seconds but let's get there, one step at a time. So our plan is to figure out what is this position*x 1*because after we know this position, we can figure out how much time it takes to get from*x 1*to*x 2*given this constant speed and we'll add that time to 7 seconds to get the total time it takes to do the race from here to the finish line and then we'll compare that with the time it takes with no acceleration. So*x 1*is going to be*x naught*plus*v naught*times*t 1*plus one-half*at 1 squared*, that's equation [2.53], and we substitute in 0 plus the initial speed of 11.5 meters per second multiplied by 7 seconds plus a half times 0.5 meters per second squared times 7 seconds squared giving 92.75 meters. So we can update our picture and write down*x 1*is 92.75 meters and the picture is the same as here but now we are considering this when the cyclist is over here. Okay. So cyclist has finished their acceleration and now they are here and then they are gonna continue on at that constant speed of 15 meters per second and we want to know how much time this will take and we can figure out that time using equation 50 because we know what*x 1*is and we can use*x 2*equals*x 1*plus*v*times*t 2*minus*t 1*; that's sort of a slight variation of how you would normally write equation 50, the way they write it in the textbook is*x*equals*x naught*plus average speed times time but this time is understood to be the elapsed time which in this case is*t 2*minus*t 1*and average speed is just going to be the speed because there is only one speed, it's constant and there we go and we are dealing with*x 2*and*x 1*on this picture instead of*x*and*x naught*so let's solve for*t 2*. First, we'll subtract*x 1*from both sides and then we'll also divide both sides by*v*and we get*t 2*minus*t 1*, after you switch the sides around, equals*x 2*minus*x 1*all over*v*and then add*t 1*to both sides to solve for*t 2*. So*t 2*is*x 2*minus*x 1*over*v*plus*t 1*. So that's the final position, 300 meters, minus the position at which acceleration ends and that's 92.75 meters all divided by the constant speed during this time interval of 15.0 meters per second and add to that the 7 seconds and we get a total time of 20.8167 seconds is the time at which the cyclist reaches the finish line after the very beginning before the acceleration started or right when it started I suppose you could say. Okay so that's the total time to finish the race given a period of acceleration. Now if there is a different scenario where the cyclist does not accelerate at all then*x 2*is going to be*x naught*plus this initial speed which is 11.5 meters per second multiplied by the time at which they pass the finish line minus the initial time but*t naught*is 0 here and this is what we wanna solve for here, this*t*constant and so we'll subtract*x naught*from both sides and then divide both sides by*v naught*and switch the sides around and we get that the time in the scenario where there's no acceleration is just the amount of distance covered 300 meters divided by the constant speed at which they are covering that distance of 11.5 meters per second for 26.0870 seconds. So the time saved then by doing acceleration is this large time where there's no acceleration or the time when it's constant speed at the initial speed in other words minus the total time when there is acceleration. So we take that difference and we find that the cyclist saves 5.27 seconds by doing that period of 7 seconds of acceleration. Okay. Part (c) says there's a competitor who begins 5 meters ahead of the initial or original cyclist and this competitor however does not accelerate and they go at a speed that is constant which is 11.8 meters per second and it turns out that this black cyclist is going to finish first and the question is how much time do they finish ahead of this blue cyclist and also what will the blue cyclist's distance from the finish line be when the black cyclist finishes? So how far ahead in time and in distance will the black cyclist finish is the question? Now I'm using a little symbol here, prime, to label all these variables to do with the blue cyclist so the initial position of the blue cyclist is not zero; normally, we take this initial position to be zero but we are not going to do that here, we are going to say that it's 5 meters because they are starting 5 meters ahead of the black cyclist and*t naught prime*will still be 0 seconds and because, you know, the clock starts ticking at the same time for both of them and*v prime*is the blue cyclist's speed and*x 2 prime*—there's no need for a prime there because it's the same finish line for both cyclists and it's 300 meters— but*t 2 prime*however will be different because this blue cyclist will take a different amount of time to finish the race and this*x 1 prime*and*t 1 prime*, they designate this position where the blue cyclist is when the black cyclist finishes. Okay. So let's figure out how long it takes the blue cyclist to finish the race and then we'll take the difference between that time versus the black cyclist's time and then that will be the amount of time by which the black cyclist is ahead of blue cyclist. So*x 2 prime*is gonna be*x naught prime*plus*v prime*times*t 2 prime*, this is equation [2.50]. Now we can solve for*t 2 prime*by subtracting*x naught prime*from both sides and then dividing both sides by*v prime*and so we have 300 meters minus 5 or in other words, 295 meters has to be covered at a constant speed of 11.8 meters per second which will take 25.0 seconds so that's the time for the blue cyclist to finish the race. And the difference in time then between the two cyclists is 25 seconds for the blue cyclist minus 20.81, which is something we calculated way up here— the total time for the accelerating cyclist— and that difference is 4.2 seconds. So the black cyclist in this picture, the one who does accelerate, finishes 4.2 seconds before this blue cyclist despite starting 5 meters behind as a result of its acceleration. Okay. So how far position-wise is this blue cyclist behind when the black cyclist finishes? Well, it's gonna be*x 2 prime*which is the finish line minus*x 1 prime*so this distance in other words that's*x 2 prime*minus*x 1 prime*and we'll figure out what*x 1 prime*is by using this equation which says the initial position plus the speed multiplied by this time*t 1 prime*and we know what this*t 1 prime*is because it's the time it takes for this cyclist to finish so this cyclist is gonna finish in in 20.8167 seconds and where will this non-accelerating cyclist be at that time. So the non-accelerating cyclist begins at 5 meters and then add to that 11.8 meters per second multiplied by this total time for the accelerating cyclist to finish and that gives 250.637 meters. So this position here,*x 1 prime*, is 250.637 and so that means they are 300 meters, which is the finish line, minus that*x 1 prime*distance which is 49 meters behind when the accelerating cyclist finishes.
## Comments

Submitted by blueFZ07 on Tue, 10/05/2021 - 19:50

??

Submitted by ShaunDychko on Wed, 10/06/2021 - 08:26

notgo 15m/s for the whole 300m. Rather, acceleration begins at 11.5 m/s at the beginning of the 300m, then arrives at 15 m/s after 7.00 s. The confusion is in the mistaken belief that the cyclist goes 15m/s constant speed for 300m.All the best,

Shaun

In reply to In finding the time saved to… by blueFZ07