Skip to main content
Ch 02: Kinematics in One Dimension

Chapter 2, Problem 2

A sprinter can accelerate with constant acceleration for 4.0 s before reaching top speed. He can run the 100 meter dash in 10.0 s. What is his speed as he crosses the finish line?

Verified Solution
Video duration:
9m
This video solution was recommended by our tutors as helpful for the problem above.
914
views
Was this helpful?

Video transcript

Hey everyone. Welcome back in this problem. A car takes off and accelerates uniformly for the 1st 7. seconds to attain its peak speed. The car covers one km in 45 seconds and were asked to determine The car's speed as it crosses a point marking one km. Alright, so let's think about this. We have to sort of stages to this problem. Okay. In stage one the car is accelerating uniform uniform acceleration. Now we know that the card takes off. So the initial speed is going to be 0m/s. We don't know the final speed and I'm going to say that this is going to subscript these with ones to indicate stage one. We don't know the final speed. We don't know the acceleration. We know it's accelerating, but we don't know the value of the acceleration. We don't know how much distance the car travels And we know that the time it spends doing this acceleration is 7.5 seconds. Now, be careful. I've said we don't know the distance, we have a distance given in the problem of one kilometer, but that happens in 45 seconds. Okay. In this 1st 7.5 seconds of acceleration, we don't know the distance we travel okay. Stage one. Now, we also have stage two. Once that car reaches its peak speed, it stops accelerating, it makes it to that maximum speed and then just continues driving. Okay. So we actually have no acceleration. Now, when we have no acceleration, we only have three variables. We have to consider, we have the speed, we're gonna call it V two. Now, this speed V two is going to be related to stage one. How well this is going to be equal to the final speed from stage one. Okay. The car accelerates, it gets to some maximum speed. That's the final speed of stage one and then it continues at that speed in stage two. Okay. So V two is going to be equal to V F one. Now we also have the distance traveled in stage two. Now the distance traveled in stage two, we don't know, but we know we can relate it to stage one. We know that the total distance we travel in both stages is one kilometer. So the distance we travel in the second stage is going to be one kilometer minus whatever distance we covered in the first stage. Now we want this in our standard unit. So we call that one kilometer is equivalent to 1000 m. Okay. We multiply by 1000 and so D two is going to be equal to 1000 m minus D one. And our time T two were told that it takes seconds. I need to cover this entire one kilometer and we spend 7.5 seconds in the first stage. And so our time spent in the second stage is going to be 45 seconds. That total time minus 7.5 seconds. The time spent in the first stage, which gives us a time of 37.5 seconds. All right. No, We have two stages. What we're trying to find Is the value of EF one, which is equivalent to me too. That's what we're trying to find. Now, using the information in stage one, we only have two known variables. We have uniform acceleration. So we can use our um equations or Kinnah Matic equations. But with two known variables, we won't be able to solve for VFR. In this second stage, we have two unknowns. We have this VF one and D one, V two and V two, but these two are related okay. We have V F one and D one in both stage one and two. And so we can write two equations with two unknowns and then sold. So we're going to choose for stage one, the equation that does not include the acceleration, Okay. And that's gonna be the following distance traveled. The one Is equal to 1/2. If he not one, the initial speed plus Vina VF one, the final speed Times The Time T one. So D one is going to be equal to one half. Our initial speed is zero. We get V F one times the time 7. seconds. And if we simplify that right hand side, we have at the distance D one is equal to 3. seconds Times the final speed VF- one. Now, we're gonna call this equation one. We can't simplify that anymore yet, but we're going to come back to it. We're gonna do the same for stage two. We're going to write as much of the equation out as we can. And because we have no acceleration, the equation we use is just the velocity is equal to the distance over time. We've used subscript two to indicate stage two here. Something in what we know, we know that the speed V two is equal to the final speed from stage one V F one. We know that the distance traveled in stage two is going to be 1000 m minus the distance traveled in D one Divided by the time 37.5 seconds. Now, we could stop here. We could substitute this quantity for VF one into equation one stall for via sol for D one. Okay. And then substitute that back into this value to get VF one. However, if we rearrange first, if we isolate the one, we write D one in terms of the F one, we can substitute D one into equation one and solve for V F one. Okay. So we can, if we do it that way, we only have one solution step versus having to substitute back into the original equation, either one will get you the same answer, the right answer. We're just trying to save a little bit of calculation work. All right. So let's rearrange this guy first. So we multiply by 37.5 seconds, we have 37.5 seconds times V F one is equal to 1000 m minus the distance D one which tells us that the distance D one is equal to 1000 m -37.5 2nd Times VF one. Okay. So we'll call this equation to and we are gonna substitute equation two into equation. What? Okay. They both have D1, they both have the F 1, 2 equations, two unknowns. Alright. So if we substitute D one is equal to 1000 m minus 37.5 seconds times VF one, we're going to get 1000 m -37.5 2nd Times VF one is equal to 3.75 seconds times via one we want to solve for V F one. So we're gonna move this 37.5 seconds times V F one to the left hand side or sorry to the right hand side. We're gonna have 1000 m is equal to V F one times 37.5 seconds plus 3.75 seconds. One final step. We need to divide, we get that V F one is equal to 1000 m over 41.25 seconds. If we work this out, we get a speed of 24.24 repeated seconds. Whoops. Not seconds. Meters per second. There we go. And that is the speed that we were looking for. We go back up to our answer choices. That was this car speed when it made it to that one km mark. Okay. That is going to correspond with answer choice. A 24.2 m/s. Thanks everyone for watching. I hope this video helped see you in the next one.
Related Practice
Textbook Question
FIGURE EX1.18 shows the motion diagram of a drag racer. The camera took one frame every 2 s. (b) Make a position-versus-time graph for the drag racer. Because you have data only at certain instants, your graph should consist of dots that are not connected together.
289
views
Textbook Question

FIGURE P1.58 shows a motion diagram of a car traveling down a street. The camera took one frame every 10 s. A distance scale is provided.


<IMAGE>


b. Make a position-versus-time graph for the car. Because you have data only at certain instants of time, your graph should consist of dots that are not connected together.

84
views
Textbook Question
A motorist is driving at 20 m/s when she sees that a traffic light 200 m ahead has just turned red. She knows that this light stays red for 15 s, and she wants to reach the light just as it turns green again. It takes her 1.0 s to step on the brakes and begin slowing. What is her speed as she reaches the light at the instant it turns green?
819
views
Textbook Question
A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the jet is moving with a speed of 400 m/s. What is the magnitude of the jet's acceleration, assuming it to be a constant acceleration?
1085
views
Textbook Question
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 m/s². a. How much distance is between you and the deer when you come to a stop?
2699
views
Textbook Question
A car starts from rest at a stop sign. It accelerates at 4.0 m/s² for 6.0 s, coasts for 2.0 s, and then slows down at a rate of 3.0 m/s² for the next stop sign. How far apart are the stop signs?
1582
views