Asked by sh
If a ball is thrown into the air with a velocity of 30m/s, its height in metres after t seconds is given by y=30t-4.9t^2
Fing the average velocity for the time period beginning when t=2 and lasting 1second.
I did V(1)=lim h->0 [[30(1+h)-4.9(1+h)^2]-[30(1)-4.9(1)^2]]/h
After I expanded everything, I am still unable to cancel the h on the bottom.
Thanks in advance.
Fing the average velocity for the time period beginning when t=2 and lasting 1second.
I did V(1)=lim h->0 [[30(1+h)-4.9(1+h)^2]-[30(1)-4.9(1)^2]]/h
After I expanded everything, I am still unable to cancel the h on the bottom.
Thanks in advance.
Answers
Answered by
Reiny
Note that it said "average velocity", what you tried to do is the instantaneous velocity at a time of t = 1 sec
(it said, "lasting 1 second", meaning there was a duration of 1 second)
so you want the average velocity between 2 and 3 seconds.
height at 2 sec = 30(2) - 4.9(4) = 40.4
height at 3 sec = 30(3) - 4.9(9) = 45.9
so average velocity = (45.9-40.4)/(3-2) = 5.5 m/s
btw, when you expanded your expression you should have arrived at
lim [30 + 30h - 4.9 - 9.8h - 4.9h^2 - 30 + 4.9]/h
= lim[20.2h - 4.9h^2]/h
= lim [20.2 - 4.9h] as h ---> 0
= 20.2
which would be the instantaneous vel at t=1
(it said, "lasting 1 second", meaning there was a duration of 1 second)
so you want the average velocity between 2 and 3 seconds.
height at 2 sec = 30(2) - 4.9(4) = 40.4
height at 3 sec = 30(3) - 4.9(9) = 45.9
so average velocity = (45.9-40.4)/(3-2) = 5.5 m/s
btw, when you expanded your expression you should have arrived at
lim [30 + 30h - 4.9 - 9.8h - 4.9h^2 - 30 + 4.9]/h
= lim[20.2h - 4.9h^2]/h
= lim [20.2 - 4.9h] as h ---> 0
= 20.2
which would be the instantaneous vel at t=1