For an object starting from rest and accelerating with constan...

Question

For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of the time. If an object travels 2.0 furlongs in the first 2.0 s, how far will it travel in the first 4.0 s?

Accepted Answer

Hey, everyone in this problem, we're told that kinematic shows if a motorcycle starts from rest and accelerates uniformly, the distance covered is proportional to the square of the change in time. In the first three seconds. A motorcycle covers 12 m. We're asked to determine the distance covered by the motorcycle in the first eight seconds. The answer traces were given are a 32 m. B 85 m C 1.7 m and D 380 m. Now this is a motion problem. OK? And we're told that we have uniform acceleration, which means that we're gonna be using our U AM equations or our kinematic equations. If that's what you'd like to call them, we have to be careful here. OK? If we just consider one set of variables for the eight second time period, we're trying to figure out the only information we really have for that period is a time. OK? The distance we're told about is only for the first three seconds. And the initial speed we're given is from the first from from time zero. So we have that initial speed and the 12 m, but only for the three second time period. So we're gonna break this up into two stages. So the first is gonna be stage one and this is gonna be that first three, second time period that we have information about the distance child. So our initial speed VNAT one is going to be 0 m per second because we start from rest and we don't know what the final speed of this stage will be. We don't know what the acceleration will be. We're told that there is uniform acceleration but we don't know what that will be. We know that the distance traveled is 12 m and that the time taken is three seconds. So this is our first three seconds. Then we're gonna move on to stage two. And in stage two, our initial speed is gonna be what? Well, this initial speed is gonna be the final speed from stage one K stage one takes us from zero seconds up to three seconds. Stage two is gonna start at that three seconds and end at the eight seconds. Ok? So our initial speed in stage two is just gonna be the final speed from stage one. We don't know what our final speed will be. The acceleration A two. We're told that we accelerate uniformly. So the acceleration is gonna be the same as the acceleration A one. We don't know the distance D two, but we need to find that out in order to answer the question. OK. We're looking for the distance covered by the motorcycle in the first eight seconds and the time T two. Ok. This is gonna be the total time taken eight seconds minus the time we spend in the first stage three seconds. Ok, which gives us a time, T two of five seconds. So we have these two different stages of motion and we're gonna start with stage one. Ok? We have three known values in stage one. We know the initial speed, we know the distance and we know the time, that means that we can find one of the other variables. And we're gonna choose to solve for the acceleration A one first. So that we can use that in our second equation. When we solve for a one, we're still gonna have only two notes. So we're also gonna have to solve for VF one. OK. So let's start with a one. We're gonna choose the kinematic equation that does not include the final speed. And we have that D one is equal to the initial velocity V, not one multiplied by the time T one, what's one half times the acceleration? A one multiplied by the time T one squared substituting in the values that we have, we have 12 m is equal to V 01 is 0 m per second. So this entire term V 01 multiplied by T one goes to zero. We have one half multiplied by the acceleration. A one multiplied by the time T one which is three seconds squared. So we have 12 m simplifying on the right hand side, three squared gives us nine times a half is 4.5 name the unit of second squared times the acceleration A one and the soft for a one, we divide by 4.5 seconds squared. And we get that A one is equal to 2.66 repeated meters per second squared. All right. So we have this acceleration a one, I'm gonna put it in a box up here and then write it in the information we have for stage two. Because now we know this acceleration in stage two is gonna be the same as the acceleration in stage one, which is 2.66 repeated meters per second squared. All right, we're gonna do the same thing. Now, in order to find VF one, we can find VF one, then we know the initial speed for stage two, then we'll have the initial speed, the acceleration and the time that's three known values. And we'll be able to solve for the distance D two that we're looking for. So we're gonna choose an equation that includes VF one this time, we're just gonna choose the simplest one to calculate because we know all of the other variables we have VF one and S equal to V 01 plus a one, T one we're trying to solve for VF one. We know that V knot is just zero VK 01 is 0 m per second. So we have the acceleration which we just found to be 2.66 repeated meters per second squared multiplied by the time of three seconds for speed of 8 m per second. And, and so that final speed VF one is going to be 8 m per second, not 8 m per 8 to 8 m per second. There we go. And we're gonna put this one in green. And we're gonna also include this in our information for stage two. So V not two, which is VF one is going to be eight meters per second. All right. So we have three knowns in stage two. Now, we can find this value of D two that we're looking for. We're gonna choose the equation that doesn't include VF two because we don't have information about VF two. And that's not what we're looking to find. And that equation is gonna be the same as the equation we used for stage one to find the acceleration. So we have D two is equal to V 02, multiplied by T two plus one half, multiplied by A two multiplied by T two squared. Substituting in the information, we know we have 8 m per second times five seconds. Remember this is stage two. So our time is now five seconds, one half multiplied by 2.66 repeated meters per second squared multiplied by five seconds squared. And if we work out the right hand side, let's look at the units first, we have meters per second time second, which gives us a unit of meter in this first term, then we have meters per second squared times second squared, which also gives us meters in the second term. So meters plus meters, we have a unit of meters and we get 73.3 repeated meters for our distance D two. So we have our distance D two, be careful. We found a distance. The question is asking for a distance. And so it can be really easy to stop at this point and say that's our answer 73.3 m. But remember we're looking for the distance covered in the first eight seconds. The distance we have found here is just the distance for the second stage. So we need to add the distance we traveled in the first stage as well. And so the total distance covered in the first eight seconds, total distance. Oh and first eight seconds is gonna be the sum of the two distances D one plus D two. And the distance covered in stage one was 12 m. The distance covered in stage two we found was 73.3 repeated meters. And this gives us a total distance in the first eight seconds of 85.3 repeated meters. And that is a distance that we were looking for we can go back up to our inter traces now and compare what we found. And we found that the distance covered in the first eight seconds was 85.3 m. So if we approximate to the nearest meter, we see that this corresponds with answer trace b 85 m. That's it for this video. Thanks everyone for watching. See you in the next one.

For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of the time. If an object travels 2.0 furlongs in the first 2.0 s, how far will it travel in the first 4.0 s?

Key Concepts

Constant Acceleration

Kinematic Equations

Proportionality in Motion

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