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Ch 02: Motion Along a Straight Line

Chapter 2, Problem 2

A car is stopped at a traffic light. It then travels along a straight road such that its distance from the light is given by x(t) = bt2 − ct3, where b = 2.40 m/s2 and c = 0.120 m/s3. (a) Calculate the average velocity of the car for the time interval t = 0 to t = 10.0 s.

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Welcome back everybody. We have an ambulance whose position with respect to time is given by this equation here where B is 3.1 and C is .147. So I'm gonna go ahead and plug those in that we get that. The position with respect to time is equal to 3.10 times t squared minus 0. t cubed. And we are asked to find the average velocity over this interval of time. Well, the formula for average velocity is given by our change in position over our change in time. Well, our change in time is going to be seven seconds minus zero seconds. So that's just gonna be seven on the bottom. But we still need our change in position or our final position minus our initial position. Let's go ahead and find that by plugging in our different values of T. X evaluated at zero seconds is going to be 3.10. I'm zero squared minus 0. times zero. Cute course this is just zero. Now x evaluated at seven, It's equal to 3.10. I'm seven squared -0.147. I'm seven cubed. Which when you plug into your calculator you get 10, Now that we have. Initial and final position, let's go ahead and plug it into this formula right here. Just once more as a reminder, our average velocity ancient X. Or changing t Equal 10, 1.479 0 over seven. Which when you plug into your clock calculator, you get 14. m/s, which corresponds to our answer choice B. Thank you guys so much for watching. Hope this video helped, and we will see you all in the next one.