Suppose the position of an object moving horizontally along a ...

Question

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.
Determine the velocity and acceleration of the object at t = 1.
f(t) = 2t³ - 21t² + 60t; 0 ≤ t ≤ 6

Determine the velocity and acceleration of the object at t = 1.

f(t) = 2t³ - 21t² + 60t; 0 ≤ t ≤ 6

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So, first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. The position of a car moving in a straight line is given by the functions of T is equal to T 3 minus 9 T 2 plus 24 T, where S is in kilometers and T is in hours. Find the car's velocity and acceleration at T equals 4. Awesome, so that's our end goal. Our end goals we're asked to solve for two separate answers. We're trying to figure out the car's velocity, that's our first answer, and the car's acceleration, our second answer, both at T equals 4. So with that said, let's read off our multiple choice answers to see what our final answer pair might be, noting that the velocity is going to be given in units of kilometers per hour, and our acceleration value is going to be in units of kilometers per hour squared. So A is 0 and 6, B is 8 and 6, C is 0 and 12, and D is 8 and 12. Awesome. So first off, what we're trying to solve for is we're ultimately trying to figure out for the first part is we're trying to figure out the car's velocity at T equals 4. So let's make a note of this. So the velocity and let's call it V for velocity at T equals. 4. That's our first step. So we need to define our position function first and let us know that our position function is given to us as SFT is equal to T power of 3 minus 9 T 2 plus 24 T. Fantastic. So now our next step is we need to determine the velocity function VFT, and in order to do that, we need to recall and note that the velocity function VFT is the first derivative of the position function SFT. So therefore, VFT. is equal to DS of T by DT and this is equal to 3T2 minus 18 T + 24. Fantastic. So now, our next step is we need to substitute in the value of T equals 4 into our velocity function, and when we do that, we will find that V of 4. is equal to 3 multiplied by 4 to the power of 2 minus 18 multiplied by 4 plus 24 will be equal to 0 kilometers per hour. And once again, we're just plugging in the value of 4 in for T. Awesome. So that will mean our first answer is 0 kilometers per hour. Hooray, we did it. We found our first answer. So now we need to solve for our second answer, which is the acceleration of the car at T equals 4. So now, let's make a note of that. So now we're solving for the acceleration, which we'll call capital A, or rather we'll just call it lowercase a. So now we're trying to find the acceleration A at T equals 4. That's our next answer we're trying to solve for. And in order to solve for this, we need to recall and note that the acceleration function A T is the first derivative of the velocity function via T. So that means that AF T is equal to DV of T, so DV of T by DT, which is equal to 6 T minus 18. Fantastic. So now, our next step. we need to substitute in our value of T equals 4 into our acceleration function. So we need to say that A of 4, so A of 4, since we're plugging in 4 for T, so A of 4 is equal to 6 multiplied by 4 minus 18, which will be equal to 6 kilometers per hour squared. And that is our final answer for the acceleration. Hooray, we did it. We found both of our final answers. And as we could see, looking at our multiple choice answers, the final answer has to be multiple choice answer letter A, which once again is velocity is 0 kilometers per hour, and the acceleration is 6 kilometers per hour squared. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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