Respuesta :
Before you go through reading this, I came to the conclusion that the height was 1.5 ft off the ground (the answer yay now no looking!)
Circumference of tire helps with revolutions.
C=2pi*r
45rpm (revolutions per minute)
Diameter is given, so adjust circumference formula accordingly.
C=pi*d
3pi=9.425 (approximately)
So 9.425 ft is the circumference of the tire, and it revolves 45 times per minute. In order to answer the question you need to know how fast it is going so take one revolution 9.425 and multiply by 45 revolutions. This gets 424.11501
Every minute that's how much ft it goes through. (424.1150082346221)
Assuming t=0 means the point on the bottom of the tire touching the ground (hence height 0) then it could have a possibility of a height of 3 ft above the ground (max, cus that's the diameter), now what id do is Id divide by 60 and multiply by 45 as it asks for "45 seconds" or you could divide by 4 multiply by 3 (60/4=15, 15*3=45..)
424.1150082346221 divided by 60 is 7.068583470577035 (ft per second)
So 7.068583470577035 times 45 is 318.0862561759666 ft
After 45 seconds, 318.0862561759666 ft has been gone through. ( turns out maybe you don't need the speed?? *sigh of relief* )
Hmm now what..
...
*just thought for 10 mins cus I learned this last semester and is a scrub*
K now I'd divide that number by the circumference.. I think..
I did that and got 33.75. So you know it went through 33 revolutions, now you ignore the 33.
You ignore this because each revolution it completed, the point's height returned to its original (0). Now take the circumference and multiply by .75 to get how much of the circumference was completed in this last revolution. Nvm don't do that last sentence that was like useless :D. K so it got .75 of the circumference, so that means you can think of a unit circle and in the circle imagine starting from the bottom where the point is and following around it to complete .75 of it, this'll leave 1/4 unshaded for example. In a unit circle you'd have (1,0) (right side of the circle) if .75 of it was gone around. So from the points perspective (saying it was height 0 again), the tire stopped halfway through (so the point on the unit circle (1,0) relative to the point at h=0 would be like resetting the y-axis halfway down the rest of the tire, so the point's height is (1,0) unit circle but actually like (1,1.5 ft) [i just screwed up the idea of the unit circle passing 1 /-:]
Anyways I conclude that the point ends at 1.5 ft above the ground. Your welcome you probably didn't understand a lot because I confused myself a lot and used tons of unneeded information!!!
This might not be right but it was my thought process, so if it's wrong I'm sorry :-(
And if it was wrong then atleast it led you in the right direction :-)
Circumference of tire helps with revolutions.
C=2pi*r
45rpm (revolutions per minute)
Diameter is given, so adjust circumference formula accordingly.
C=pi*d
3pi=9.425 (approximately)
So 9.425 ft is the circumference of the tire, and it revolves 45 times per minute. In order to answer the question you need to know how fast it is going so take one revolution 9.425 and multiply by 45 revolutions. This gets 424.11501
Every minute that's how much ft it goes through. (424.1150082346221)
Assuming t=0 means the point on the bottom of the tire touching the ground (hence height 0) then it could have a possibility of a height of 3 ft above the ground (max, cus that's the diameter), now what id do is Id divide by 60 and multiply by 45 as it asks for "45 seconds" or you could divide by 4 multiply by 3 (60/4=15, 15*3=45..)
424.1150082346221 divided by 60 is 7.068583470577035 (ft per second)
So 7.068583470577035 times 45 is 318.0862561759666 ft
After 45 seconds, 318.0862561759666 ft has been gone through. ( turns out maybe you don't need the speed?? *sigh of relief* )
Hmm now what..
...
*just thought for 10 mins cus I learned this last semester and is a scrub*
K now I'd divide that number by the circumference.. I think..
I did that and got 33.75. So you know it went through 33 revolutions, now you ignore the 33.
You ignore this because each revolution it completed, the point's height returned to its original (0). Now take the circumference and multiply by .75 to get how much of the circumference was completed in this last revolution. Nvm don't do that last sentence that was like useless :D. K so it got .75 of the circumference, so that means you can think of a unit circle and in the circle imagine starting from the bottom where the point is and following around it to complete .75 of it, this'll leave 1/4 unshaded for example. In a unit circle you'd have (1,0) (right side of the circle) if .75 of it was gone around. So from the points perspective (saying it was height 0 again), the tire stopped halfway through (so the point on the unit circle (1,0) relative to the point at h=0 would be like resetting the y-axis halfway down the rest of the tire, so the point's height is (1,0) unit circle but actually like (1,1.5 ft) [i just screwed up the idea of the unit circle passing 1 /-:]
Anyways I conclude that the point ends at 1.5 ft above the ground. Your welcome you probably didn't understand a lot because I confused myself a lot and used tons of unneeded information!!!
This might not be right but it was my thought process, so if it's wrong I'm sorry :-(
And if it was wrong then atleast it led you in the right direction :-)
