Required information Problem 11.004 - Railroad car NOTE:...

Required information Problem 11.004 - Railroad car NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. A loaded railroad car is rolling at a constant velocity when it couples with a spring and dashpot bumper system. After the coupling, the motion of the car is defined by the relation x = 60e^-4.8tsin16t, where x and t are expressed in millimeters and seconds, respectively. Problem 11.004.b - Railroad car Determine the position, velocity, and acceleration of the railroad car when t = 0.7 s. The position, velocity, and acceleration of the railroad car when t = 0.7 s are 2.07 mm, 12.88 mm/s, and 453.426 mm/s^2, respectively.

Transcript

00:01 All right, so we're told that the position in millimeters is given as a function of time in seconds by this function here.

00:09 So if we wanted the velocity, it's going to be the derivative of the position function.

00:15 So using the product rule, since we have an exponential multiplied by a trigonometric function, first let's take the derivative of this one.

00:22 So it's going to be 60 times negative 4 .8e to the negative 4 .8t, and then we keep the sign of 16t plus now keep the first one.

00:38 So we have 60e to the negative 4 .8t multiplied by the derivative, which is cosine of 16t times 16.

00:48 So 16 cosine of 16 t, which we can go ahead and simplify because we're going to have to take the derivative again for acceleration.

00:57 So we might as well.

01:01 So we have negative 4 .8 times 60 is negative 288 times e to the negative 4 .8 t sign 16t.

01:21 And then the 16 and the 60 we can multiply together.

01:25 And we get 960, e to the negative 4 .8t cosine of 16t.

01:35 All right.

01:35 And then the acceleration, we take the derivative of a velocity.

01:39 So again, applying the product rule.

01:41 So negative 288 times negative 4 .8e to the negative 4 .8 t, sign of 16t.

01:53 So that takes care of taking the derivative of the first term, plus keep that first exponential term and take the derivative of the second term.

02:04 So it's going to be 16 cosine 16t.

02:08 Plus now handling this second term here.

02:13 Derivative of the exponential first plus 960e, leave the exponent intact.

02:29 And then we have negative 16 sign of 16t because the derivative of cosine is negative sign.

02:39 All right.

02:40 And then we don't have to, well, we might as well.

02:46 Negative 288 times negative 4 .8.

02:49 The negative signs cancel out.

02:50 And this becomes just 100 and 1 ,382 .4.

02:56 As the exponent here.

02:59 And then 280 times 16 is 4 ,608.

03:05 So i'll just erase that and that.

03:15 Negative 4 ,608.

03:20 Okay, 960 times 4 .8, negative.

03:27 Oops.

03:33 Oh, i can just cross it out, actually.

03:38 Negative 4 ,608.

03:41 And then 960 times 60.

03:52 This is negative 15 ,360.

03:59 All right.

04:00 So now, if we want the position at times t equals 0 .7, so x of 0 .7 is going to be 660e, so the negative 4 .8 times 0 .7 times the sign of 16 times 0 .7...

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