A dog running in an open field has components of velocity υx = 2....

Question

A dog running in an open field has components of velocity υx = 2.6 m/s and υy = −1.8 m/s at t1 = 10.0 s. For the time interval from t1 = 10.0 s to t2 = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s2 and direction 31.0° measured from the +x–axis toward the +y–axis. At t2 = 20.0 s, (a) what are the x- and y-components of the dog’s velocity?

Accepted Answer

Welcome back everybody. We are looking at a football field top down and we have an interval of time from eight seconds to 15 seconds. Now, we are told that a football player at the very beginning as an initial horizontal velocity of negative three m per second. And we are also told that he has an initial vertical velocity Of 1.6 m/s. Now, during this interval of time, we are told that he is running with an acceleration of 0.6 m per second squared. And we are told that this acceleration is at an angle of 40 degrees above the negative X axis. And we are tasked with finding what is going to be his final X and Y velocities are final X. We need to find and our final y we need to find this is a lot of information to absorb. But I'm gonna make this easy. We're simply going to use this Kinnah Matic equation that our final velocity is equal to our initial velocity plus our acceleration times the change in time. And we're just gonna do this for the X components and the Y components. But what are the X and Y components? So let's go and find those first. The initial velocity in the X direction while we're given that is negative three m per second or delta T. Well, that's just gonna be how much time has passed. So that's gonna be 15 minus eight seconds. That is just seven seconds. Now, what is our acceleration in the X direction? Well, we are looking for this vector right here. So this is going to be equal to the magnitude of our acceleration factor times the cosine of our angle and it is going to be negative since our vector is pointing left. Great. So now let's find our Y components here. So our initial vertical velocity were given as 1.6 m per second. Are delta T. Does not change. That is going to stay at seven seconds. Now we need our vertical acceleration. Well this time we are looking for this vector right here. So that vector is going to be R a Y r a way. This time is going to be our magnitude 0.6 times the sine of 40 and it is positive since this is in the positive Y direction. Now that we have all the information we need, let's go ahead and find our final X and Y components of our velocity. So our final X component for our velocity equal to the initial X component of our velocity plus the X component of our acceleration times our delta T. So plugging in the values here, we get negative three plus negative 0.6 times the cosine of 40. All times seven squared. Which when you plug into your calculator you get negative 6.2 m per second. Now our final Y velocity same similar thing here. So it's equal to our initial Y velocity plus the Y component of our acceleration times the change in time plugging values here, we have 1.6 plus 0.6 times the sine of 40 all times seven squared which when you plug into your calculator, you get 4.3 m per second. So we have now found our final X component over velocity and the final Y component of our velocity corresponding to answer choice. Thank you guys so much for watching. Hope this video helped. We will see you all in the next one.

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