If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:6:43

Video transcript

problem number one at time T a particle moving in the XY plane is at position X of T Y of T where X of T and y of T are not explicitly given for T is greater than or equal to zero the derivative of X with respect to T is for T plus one the derivative of Y with respect of T is sine of T squared and at time T equals zero X of zero zero and Y of zero is negative four fair enough let's do Part A find the speed of the particle at time T equals three and find the acceleration vector of the particle at time T is equal to three as well so the speed is really just the magnitude of the velocity vector so what's the velocity vector our velocity vector as a function of time is going to be equal to R the derivative of our X position as a function of time or we could say the velocity in the X Direction times the I unit vector plus the velocity in the Y Direction times the J unit vector I just wrote this in engineering notation obviously there are many ways and you could specify a vector and in this case X prime of T they've already given it to us X prime of T is 4 T plus 1 I just sorted is DX DT so this over here is 4 T plus 1 times the I unit vector and then this here this Y prime of T they gave it to us the derivative of Y with respect to T sine of T squared so plus sine of T squared plus sine of T squared times the J unit vector that is our velocity vector so what is our velocity or this is this is our velocity vector as a function of time so what is our velocity vector at time T equals 3 well we just have to substitute 3 for T so 4 times 3 plus 1 is 13 I plus 3 squared 3 squared is 9 sine of 9 plus sine of 9 j our speed what they what they want us to figure out is going to be equal to is going to be equal to or maybe I should write it as our speed at time three our speed is going to be equal to the magnitude of our velocity vector at time three which is equal to the magnitude of 13 I plus sine of nine J and then this we literally this is we use essentially the Pythagorean theorem to find the magnitude this is going to be equal to the square root square root of 169 this 13 squared plus sine of nine sine of 9 squares let me get my ti-85 out this part of the exam we are allowed to use calculators so we get the square root of 169 plus the square root of 169 plus sine of 9 sine of 9 squared alright that's 13 squared plus sine of 9 squared and we get we get 13.00 I'll just round it to 7 just to show that if we didn't get just straight-up 13 that the sine of 9 squared did do something so 13 point zero zero seven so this is approximately equal to 13 point zero zero seven so we've answered the first part the speed of the particle at time three equals zero is 13 point zero zero seven then they want us to find the acceleration vector of the particle at time T equals three so the acceleration vector is a function of time the acceleration vector as a function of time it's just going to be the second derivative of X with respect to T or you say the acceleration in the X direction as a function of time times I plus the acceleration in the Y direction so it's the second derivative of Y with respect to T in the J direction and this of course is going to be equal to the derivative of X of the derivative of X the second derivative of X of T is just the derivative of the first derivative so the first rivet is 4 T plus 1 you take the derivative of that you just get four so we get four i plus and then the derivative the second derivative or the second derivative of y with respect to t is just the derivative of the first derivative the derivative the first derivative is sine of t squared so the derivative of that is going to be 2t x times cosine of t squared and of course we have RJ we have RJ right over there and I just did this from the chain rule the derivative of T the derivative of T squared with respect to T is 2t derivative of sine of T squared with respect to T squared is cosine of T squared and so that is the acceleration vector as a function of T but they care at time T is equal to 3 so 4 still stays 4 then you have 2 times 3 plus cosine of 90 right this is going to be so acceleration vector our acceleration vector at time 3 is going to be 4 I 4 I plus and let's just put if this is this is 2 times 3 which is 6 times cosine 6 times cosine of 9 J let's get the calculator out so we actually get a value there since we they allow us to use they're almost want us to use the calculator here so 6 times cosine of 9 6 times cosine of 9 is negative 5.4 seven or five point four six seven if we want so plus negative what I'm doing this is J so plus negative five point four six seven J so that is our acceleration vector and once again I wrote it in engineering notation which I'm sure the the advanced placement exam people wouldn't mind but you could also write it in kind of in kind of your the the parametrized vector notation or your ordered pair notation as a vector you could say that the vector the acceleration vector at time three you could write it this way to for negative five point four six seven these are really specifying the same things is the X component of the vector this is the Y component of the vector this is the X component of the vector this is the y component of the vector
AP® is a registered trademark of the College Board, which has not reviewed this resource.