I really need help with this problem
an automobile accelerates from rest at 1 + 3(square root)t mph/sec for 9 seconds.
How far does it travel in those nine seconds?
I have found that the velocity, after nine seconds, is 63
I have figuredout the answer but i don't know how to do it the answer is 344.52 feet
the derivative of distance gives us velocity, and the derivative of velocity gives us acceleration.
So lets works backwards
if a(t)=1 + 3√t then
v(t) = t + 2t^(3/2) + c
but we were told that when t=0, v=0 (from rest), so c=0
Your answer of 63 is correct
now let d be the distance
then d(t) = .5t^2 + (4/5)t^(5/2) + k
but when t=0, d=0 so k=0
I think the problem is the units.
The original "a" was in mph/sec
If I recall my archaic measurement units, 1 mile was 5280 feet, and of course 1 hour = 3600 seconds
so 1mph = 5280/3600 ft/sec
so if we change our distance formula to read
d(t) = (528/360)(.5t^2 + (4/5)t^(5/2)
we can input seconds and get feet as output
when t=9
d(9) = 528/360*(.5(81) + .8(9)^2.5_
= 344.52 feet
thanks so much I was lost on that the measurement was what threw me off.
This was asked close to the year I was born
No problem! I'm glad I could help you understand the problem. Sometimes the units can be a bit tricky, but it's important to make sure they are consistent throughout the problem. If you have any more questions, feel free to ask!
You're welcome! I'm glad I could help you understand the problem. It's important to pay attention to the units in any physics problem because they can make a big difference in the calculation. In this case, the acceleration was given in mph/sec, so we needed to convert it to ft/sec to ensure that our final answer was in feet.
To convert from mph to ft/sec, remember that 1 mile is equal to 5280 feet and 1 hour is equal to 3600 seconds. So, 1 mph is equal to 5280/3600 ft/sec. By multiplying the original distance formula by this conversion factor, we ensured that our answer was in feet.
Keep in mind that understanding and managing the units is an important skill in physics. It helps you make sure that your calculations are consistent and your final answer is in the correct units. If you have any more questions, feel free to ask!