A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v = [5.00 m/s − (0.0180 m/s3)t2]î + [2.00 m/s + (0.550 m/s2)t]ĵ. (b) What are the magnitude and direction of the car's velocity at t = 8.00 s? (b) What are the magnitude and direction of the car's acceleration at t = 8.00 s?

Video transcript

Hey everyone in this problem. We have a drone flying around. Okay. It's flying over the waters of a local lagoon and we're given the velocity of a drone. Okay. As a function of time described by the equation. V T is equal to 12 m per second plus 0.9 to eight m per second cube times t squared I hat plus 20.5 m per second minus 1.35 m per second cube t squared. J had. Okay, what we're asked to do is we're asked to find the magnitude and direction of the drones, velocity and acceleration three seconds after the launch. Okay, Alright, well we're given an equation for the velocity of the drone. So let's start there. Okay, so we have V T. Okay. And what we want to find is the velocity and acceleration in three seconds. Okay, so let's go ahead and find V At three seconds. Okay, Well this is gonna be 12 m/s plus 0. meters per second. Cute times three seconds squared. I hat plus 20. m per second minus 1.35 m per second cubed times t squared which is three seconds squared. J hat. Okay, Alright, so simplifying all of this. We're going to have, is that the velocity at three seconds is equal to 12. 352. Okay. We have units meters per second and then we have meters per second cubed times second square. Okay, that's going to be meters per second as well. So we add meters per second to meters per second. So our unit is meters per second. Okay? I hat plus the same for the J Hot direction. 11.185 m per second. Okay. Alright. So we've gone ahead, substituted our three seconds and we find this equation. Now let's interpret again, we're looking for the magnitude and direction. Okay? So when we're talking and I'm gonna scroll right up here when we're talking I hat and J hat these indicate direction. Okay, So when we're talking we're talking in the X direction. This is the unit vector in the extraction. And we're talking about why we're talking the unit vector in the Y direction. Ok, so we can think of this As having velocity going to the right 12.8352 m/s. Okay. And going up, We're having an upwards velocity of 11.185 m/s. Okay, So again, I had in the extraction RJ had in the Y direction. And V three. Does that have partners here? Okay. Alright, so let's go ahead and find that magnitude. We have a right angle triangle. Go ahead and use pythagorean theorem. Okay, So V 73 squared is equal to 11. Squared times or plus 12.8352 squared. This is gonna give V3 squared is equal to 0.8466. And here we have meters per second. meters per second. Okay, we're squaring. So we get meters squared per second squared. Taking the square root we get V three is equal to positive or negative, 17.0 to 5. The square root of meters squared per second squared. We're just going to get the unit of meters per second, which is what we want for velocity. So that's good. The units check out. Okay. And we're asked for the magnitude. So just taking the magnitude, we're going to get the positive route. So we get 17.25 m per second. Okay. Alright. So we found the magnitude of the velocity. Were also asked to find the direction. Okay, So what we want to find is the state of value here. Okay, well how do we do that? Well, we know that in theater relates the opposite and the adjacent side. Okay, So we're gonna have 10 data easily to 11. m per second divided by 12.8352 m per second. Okay, so the units will cancel there. We get theta is equal to universe tangent of 11.185 divided by 12.8352 which gives data is equal to 41 .07°. Okay. Alright, so now we've found the magnitude and direction of the velocity. Okay, so we have a magnitude 17.025m/s in the direction of 41.07°. Now we need to do the same for the acceleration. Okay. All right, well let's think about this. Okay, we're given an equation for velocity. Were not given an equation for acceleration. Okay, well how can we relate velocity and acceleration? So let's go down here and I'm just going to rewrite that equation for velocity so that we have it to refer to. The equation we were given is V T is equal to 12 m per second plus 0.928 meters per second. Cute. T squared my hat plus 20.5 m/s -1.035 m/s cubed T squared J. Alright, so we have velocity. How can we relate velocity, acceleration, acceleration. We'll call it a T. Okay, it's simply the change in velocity over time. Okay, change in velocity over the change in time. So we have a D V D d v D T. Okay, Alright, so let's go ahead and take the derivative of this velocity equation. Okay, so in the I hat direction we have 12. Well, the derivative of a constant is just gonna be zero. Okay, so this term we don't have to worry about and then we have 0.928 times t squared. Ok, well, derivative of t squared the exponent two is going to come down in front? We're going to have two times 0.9 to eight. Okay. And then we're just gonna be left with T in our unit here going to be we have meters per second cubed here still. Okay, I have same thing with the jihad equation. K 20.5 derivative constant is going to be zero and the same rule applies here to the T squared. Ok. The two is gonna come in front and multiply And we're left with just a t reduced the experiment by one J. Hat. Alright so now we have the general equation for the acceleration. Okay? And we want to find the acceleration at three seconds. So let's do the same as for velocity we substitute T equals three. So for acceleration A three. Well this is gonna be two times 0.0928 times three. I have hm plus minus two times 1. times three seconds J hat And again our units here are meters per second cubed. Then we're multiplying by T. Which is a unit of second. That leaves us with meters per second squared. Okay So simplifying we get 0. I a hat. Okay And let me actually write the unit here. Meters per second squared. Ok. I had Plus and then we have negative 6.21 meters per second squared J hat. Alright so we found the acceleration. We need to find the magnitude and the direction. Okay. Alright so let's do the same for like velocity will draw a little triangle. And in this case we have I had to the unit direction. In the X direction we have a positive value. So we're going to the right mhm To the right 0.5568. And in the J hat direction we have a negative value. So that indicates going downwards KJ the unit vector in the Y direction points up. So this is going down so we're going down 6.21. Yes. We have some x component of our acceleration and AY component of our acceleration and we want to know the magnitude here. Okay, Alright, so a three squared same as with the velocity, this is a right angle triangle. We can use pythagorean theorem. So we get 0.5568 squared plus 6. squared. Okay, so a three squared is going to be 0.31 plus 5641. Okay, a three squared is equal to 38.8741. Which gives a three taking the square root is going to be plus or minus 6.235 m per second. Okay, and again we just want the magnitude. So taking the magnitude we don't need to worry about the negative and we get six point 235 m per second. And sorry, this should be meters per second squared we're talking acceleration. Okay, so we found the magnitude of the acceleration. The last thing we want to do is to find the direction. Okay, now when we're talking direction, let's remember that our angle is according to the coordinate axis. Okay, so we can find this angle here And let's just in blue call this data prime. Okay. We can find that angle there. But the angle we actually need to find is the angle that goes all the way around until that line this angle theta can we? And so in order to find data. Well we need to find 360°. Okay, the entire circle minus that value of data prime. Okay so adding those two values up data plus data Prime is going to give us 61 full circle. So we're gonna have to subtract. Okay well let's go ahead and see what we can do. So we have 360 degrees. Okay minus data prime. Now, data prime. We find it in the same way as before. Right? This is related with tan the opposite side divided by the adjacent side. So we get data, Prime is tan inverse of the ratio of those sides. 6.21 divided by 0.5568. Mhm. This is going to give us a day to prime value of 84.88°. And so we get a data value of 275.12°. And I'm gonna call this data a so we know that it's the data relating to the acceleration. Okay, so there we go. We found the acceleration and we found the direction of that acceleration. Okay, now one thing I want to note here, you might be thinking, why didn't we use the um equations? Okay, we were given a velocity equation, we could have found, you know, an initial velocity, final velocity. Um We have a time and work out the acceleration that way. Okay, Because the um equations are for uniformly accelerated motion. Okay? And in this case we don't have a constant acceleration. So those equations don't hold and that's why we can't use those equations in this case. That's why we had to take the derivative of that be equation to find a. Okay, Alright, so looking at what we found, the answer is going to be C. Okay, so the velocity is 17.5 m per second with the direction of 41.7 degrees. And the acceleration is 6.235 m per second squared with the direction of 275.12 degrees. Thanks everyone for watching. See you in the next video