Velocity at increasing vs. Decreasing rate?
- Thread starter harryyeah
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Which part of the exercise is causing you difficulty? What have you tried during your "struggles"?
Please be specific and complete. Thank you!
Hi,
I've attached a picture of a question I am struggling with. I've seen some explanations online but they are not making sense to me.
Any help would be appreciated - thank you
That part (b) is not making any sense to me. Using units of km/hr means a velocity but the problem talks about a rate of change of velocity which is km/hr2 (for this problem). From one perspective I'm getting they both start, at 7 am, at 8 km/hr and end, at 8 am, at 15 km/hr and, from another perspective, I'm getting that one was speeding up and the other was slowing down during the whole hour. Now you can't (as far as I know) start at 8 km/hr and end at 15 km/hr while slowing down the whole time.
Oh, and part a, I don't see a shaded region.
HallsofIvy
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"Lisa's velocity increases at an increasing rate and Lisa's velocity increases at an increasing rate and Kim's velocity increases at a decreasing rate" is, at best, an awkward sentence! But it is clear that both velocities are increasing so their accelerations are both positive: writing Lv and Kv for Lisa and Kim's velocities, respectively, and La and and Ka for their accelerations, we have Lv(t)= La(t)t and Kv= Ka(t)t, with La(t) and Ka(t) both positive.
Now, the fact that "Lisa's velocity increases at an increasing rate" and "Kim's velocity increases at a decreasing rate" means that La(t) is an increasing function and Ka(t) is a decreasing function. The simplest thing to do is to assume that La(t)= lt and Ka(t)= -kt with l and k both positive numbers.
Since this is posted in the "Caculus forum", I presume you can integrate those functions to find Lv(t) and Kv(t), then integrate again to find the distance run. Since they start at the same time, from the same place, and from a standstill, you can take Lv(0)= Kv(0)= 0.
For part (a) you should know:
1) For motion problems, the distance traveled is the integral of the velocity function.
2) For graphs, the integral of the function graphed is the area under the graph.
Now, the fact that "Lisa's velocity increases at an increasing rate" and "Kim's velocity increases at a decreasing rate" means that La(t) is an increasing function and Ka(t) is a decreasing function. The simplest thing to do is to assume that La(t)= lt and Ka(t)= -kt with l and k both positive numbers.
Since this is posted in the "Caculus forum", I presume you can integrate those functions to find Lv(t) and Kv(t), then integrate again to find the distance run. Since they start at the same time, from the same place, and from a standstill, you can take Lv(0)= Kv(0)= 0.
For part (a) you should know:
1) For motion problems, the distance traveled is the integral of the velocity function.
2) For graphs, the integral of the function graphed is the area under the graph.
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Thanks for your help. I do not know how to get from the acceleration function to the velocity function to the distance function. I understand how to integrate but do not know how to set up the functions in the first place to integrate them. I'm also confused how acceleration is positive in both scenarios but La(t) = lt (positive) and Ka(t) = -kt (negative). Apologies, I'm just not understanding the process to complete this question - I appreciate your help
Which part of the exercise is causing you difficulty? What have you tried during your "struggles"?
Please be specific and complete. Thank you!
Thanks for responding
I'm not sure how to use the information to generate an acceleration function (for Lisa and Kim) and in turn determine velocity and distance travelled - I can see that integration is required but I'm not sure how velocity increasing at an increasing rate vs. velocity increasing at a decreasing rate works/what it means?
Cheers!