November 30, 2020 ## An object moves with constant acceleration 4.00 m/s2

an object moves 15 meters in 3 seconds. what is its velocity (in m/s)? This is a topic that many people are looking for. voteyesons.org is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, voteyesons.org would like to introduce to you An object moves with constant acceleration 4.00 m/s2. Following along are instructions in the video below:
Course. If you havent done so yet make sure that you pause. The video video and try to answer the question on your own first before listening on in a of this question.
We are told that the objects original velocity is six meters per second. So. We can say v.
Naught is equal to positive six meters per second. And we are asked to calculate its displacement during the time interval. So displacement is typically symbolized by delta x.
And thats what were looking for we know that it has a constant acceleration of four meters per second squared. So we can say a is equal to positive four meters per second squared and then the final velocity is twelve meters per second. So we can say v.
Equals. Twelve meters per second. Weve pasted the three main equations of kinematics over here on the right hand side.
We want to make sure that we pick an equation that involves the four variables that weve listed here notice that time is not part of them so the equation that makes the most sense to use the one without time would be this one we can copy that equation down and then well simply fill in the known values some people like to fill in the values and then solve for the unknown others like to solve for the unknown and then fill in the values in our case. Well simply plug in and then solve so for v. Like we said we have 12 meters per second squared equals.
6. Squared plus 2 times 4 times. The delta x value we can square both sides.
144. And then this ones 36 4 times. 2 of course is 8.
We will then subtract 36 from both sides of this equation.

So well get 108 equals. 8. Delta x.
And then finally divide both sides of the. Equation by. 8 and we get 135 meters.
Is equal to delta x. So. This is the correct answer to part a of the question.
We go up to part b. Which says. What is the.
Distance it travels during this interval. Well we calculated the displacement as 135 meters. So you can imagine that as starting at a point.
And because it was positive displacement. It was simply moving along a straight line until it reached its final destination. Here and that displacement was thirteen point five meters.
Youll notice that because the object traveled in a straight positive direction. That the distance will also be thirteen point five meters. The only way the distance would have been different is if the particle took some sort of convoluted path in order to reach its final destination.
Something like that but of course it didnt it just traveled in a straight line. So the answer to part b. Is that the distance will also equal thirteen point five meters next on to part c it says if the original velocity was negative six meters per second.
What now is the displacement so in part c were basically going to be borrowing all of these values.

But were gonna make one variation to them because the initial velocity is now negative six meters per second. We have to make sure that we plug a negative sign in right there so we will now go ahead and use the same equation that we used before and well plug in the known values. So well have 12 and initial was now negative.
Six plus. Two times four times delta x. Basically.
When you work this out. Youre going to end up with the same value for delta x. Because when you square the negative six.
It became positive 36. So once again delta x. Would equal 13 point five meters.
So this is the correct answer to part c. Now on to part d. Which is perhaps the trickiest part of the question.
It wants the total distance that it during part c so were gonna draw a picture in order to help us understand whats happening in part c. So that can help us answer. The question for part d so remember in part c the initial velocity was negative six meters per second so we have this particle and initially its actually moving in this direction.
We know its moving to the left because the initial velocity was negative six meters per second. We also know that the final velocity from part c was positive 12 meters per second and so whats happening is this particle initially is traveling to the left with negative velocity. Its going to slow down and eventually stop then its going to need to turn around and move back until it reaches a positive final velocity of 12 meters per second.
So. What were gonna have to do is figure out how far it traveled this way along that journey and then were gonna have to find out how far it traveled this way along that journey and then add them together now its important to understand that when it reaches this point it momentarily stops so the final velocity right here is going to be zero meters per second. So lets focus our attention on the green path for now lets rewrite our known values.
The initial is negative six meters per second again.

The final is zero meters per second the acceleration was a constant believed it was four meters per second squared. Yes. And what were going to do is calculate the delta x.
There same equation as before well plug in zero. Squared. Equals.
Negative. Six squared plus two times four times delta x zero squared is zero negative. 6.
Squared is 36 2 times. 4 is 8. Lets subtract the 36 from both sides so it cancels out on the right.
And then lets divide negative 36 by 8 and we get a displacement of negative 45 meters notice the displacement is negative this should make sense in light of our diagram. If we go back up to the green path. The object was traveling overall in a leftward direction.
And so this displacement should be negative and indeed. It was now for the blue path notice for the blue path. Were now going to say the initial velocity is zero meters per second because remember it momentarily stopped at this point right here so its stopped its initial velocity is zero.
Then its going to speed up and reach 12 meters per second. So for the blue path lets list out everything we know initial is zero final is 12 acceleration is still 4. Were going to calculate delta x.
Same equation. Well plug in the final. Squared.
Equals.

The initial. Squared plus. 2.
Times. 4. Times delta x.
We get. 144 zero squared is just zero. So of course that cancels out we get eight delta x divide both sides of the equation by h.
And youre going to get 18 meters is the displacement notice thats positive displacement that again should make sense because along the blue path. The object was overall moving to the right so that displacement is positive 18 meters. Now the total distance.
Its important to understand this you dont simply add these two values together. Because distance is always a positive quantity so in other words along the green. Path even though the displacement was negative.
45 meters the distance along the green path is positive 45 meters. And of course. The distance along the blue path.
It was already positive so that remains positive 18 meters. The total distance d. Total.
Is simply the sum of these two. So you just add them together and you get 22 point. 5.
That is the final distance in part d. .