60–63. Equivalent constant velocity Consider the following vel...

Question

60–63. Equivalent constant velocity Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of ________.

v(t) = t(25−t²)^1/2, for 0≤t≤5

v(t) = t(25−t²)^1/2, for 0≤t≤5

Accepted Answer

Welcome back everyone. Given the velocity function VF T equals T2 root of 36 minus T2 for T between 0 and 6 inclusive, what constant velocity would result in the same total distance traveled over the given time period? A 8, B12, C 18, and D 24. First of all, we're going to set up the distance formula. Let's recall that we can evaluate distance by taking the integral between two time values, in this case between 0 and 6, of the absolute value of the velocity function V of T. With respect to time, so we're adding DT. Now within this time interval. T is greater than or equal to 0 but less than or equal to 6. Both t and square roots of 36 minus T2d are non-negative, meaning we can simply drop the absolute value and integrate T square root of 36 minus T2 with respect to time between 0 and 6. Let's go ahead and introduce you substitution to solve this integral. We're going to suppose that U equals 36 minus T2d. We're going to identify DU divided by DT. Which is the derivative of 36 minus T2 and that's negative 2 T. So from this relationship, we can show that T multiplied by the T, which we have within our integrat. Would be equal to DU divided by -2, or simply -1/ DU. And then we're going to change our limits of integration. The lower limit u1 would be 36 minus 0 squared, which is 36, and the upper limit U2 would be 36 minus 6 squad, which is 0. So we can now rewrite our integral as the integral from 36 up to 0 of square root. Of U multiplied by -12 DU we can factor out -12 and simply integrate square root of UDU. Applying the reverse interval property, we can state that this is 1/2 integral from 0 to 36, so we're swapping our limits of integration of you raises the power of 1/2 the U. And now let's integrate using the power rule. That's 1/2 multiplied by you raise to the power of 3 halves divided by 3 halves. Evaluated from 0 up to 36. And now what we can do is simplify. 1/2 divided by 3 halves would be 1/2 multiplied by 2/3. And that's 2 divided by 6, or it's simply 1/3. And then we have you raise the power of three halves, that's you square root of you, and we want to evaluate from 0 to 36. Let's go ahead and evaluate. That's 1/3, multiplied by 36, multiplied by square root of 36 minus 1/3, multiplied by 0 square root of 0. And this is equal to 12 multiplied by 6, which is 72. Well then, so we have identified the total distance traveled. Now, assuming constant velocity, let's recall the distance formula. Distance equals constant velocity multiplied by time. And now solving for velocity, velocity is distance divided by time. We know that our distance is 72, and our time interval between 0 and 6 is 6 minus 0, which is 6. So we get 72 divided by 6, which is 12, and this corresponds to the answer choice B. Thank you for watching.

60–63. Equivalent constant velocity Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of ________.v(t) = t(25−t²)^1/2, for 0≤t≤5

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60–63. Equivalent constant velocity Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of ________.

v(t) = t(25−t²)^1/2, for 0≤t≤5