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The position of a particle along a straight-line path is defined by $s=\left(t^{3}-6 t^{2}-15 t+7\right)$ ft, where $t$ is in seconds. Determine the total distance traveled when $t=10$ s. What are the particle's average velocity, average speed, and the instantaneous velocity and acceleration at this time?

$\begin{aligned} v_{t=10} &=165 \mathrm{ft} / \mathrm{s} \rightarrow \\ v_{a v g} &=25 \mathrm{ft} / \mathrm{s} \rightarrow \\ a &=48 \mathrm{ft} / \mathrm{s}^{2} \rightarrow \end{aligned}$

Physics 101 Mechanics

Chapter 12

Kinematics of a Particle

Motion Along a Straight Line

Newton's Laws of Motion

Applying Newton's Laws

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everyone. So what we have is the motion of a particle that is to find by this equation for position here such that the position is a function of time. Is it cubed? Minus 60 squared minus 15 T plus seven. What we're asked to solve for is the total distance traveled, the average velocity, the average speed and the instantaneous velocity at most 10 seconds. So what we're going to do first is go ahead and take our expression for s. Our position, which is to recall s, is equal to take minus 60 squared minus 15 T plus seven. And take the derivative of this. Just get our velocity. So we see that V is equal to De S DT, which gives us three D squared minus 12 T minus 15. So we can actually go ahead and solve for r instantaneous velocity here. So, by plugging in t equals 10 we'll see that the V Atiq with 10 is 1 65 ft per second. So that's one part of our answer. So we go ahead box that off next week and take the derivative of velocity to get acceleration and we can say that a is equal to DVD T, and this gives us that acceleration is a function of time. This 60 minus 12. Now we can plug in, uh, t equal to 10 and we'll see the acceleration to equal to 10 is equal to 48 feet per second squared. This is another part of our answer. We could get hood and box. That a swell. Now, given that this motion is governed by a polynomial to solve for our total distance traveled, we need to solve for when the particle change direction and also find the positive route so we can see by solving this equation for zero is that a positive route occurs when t is equal to five seconds. So therefore, what we can say that AT T equals zero s is equal to 7 ft. That he equals five s is equal to negative 93 ft and at our final time, point of T equals 10. This is equal to 257 ft. So by summing these up, we can find the total distance that has traveled. So we see that this goes from positive to negative back positive. So what this means is that we can solve for the total distance. They're saying 7 ft plus 93 ft times two because we're going down and back to 93 coming back to zero plus 257 ft. This gives us a total distance, traveled equal to 450 ft. So this will be another part of our final answer. Now, we're also asked us all for the average velocity, A T equals zero so called that average velocity is equal to Delta s all of adult T, and this will be equal to our displacement. So Delta Ass is 2 57 minus seven all over about the tea, which is 10. And this gives us average velocity 25 feet per se. And now we look at our average speed. And remember that average speed call it. The average is just total distance traveled all over Delta T. And this is equal to 450 all over 10 seconds, which gives us an average speed of 45 ft per second. And that is there a complete final answer

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