If a ball is thrown straight up into the air with an initial velocity of 85 ft/s, its height in feet after t second is given by y=85t-16t^2. Find the average velocity for the time period begining when t=1 and lasting :
.1 seconds:
.01 seconds:
.001 seconds:
Finally based on the above results, guess what the instantaneous velocity of the ball is when t=1.
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1 response so far ↓
1 Nate // Apr 11, 2008
Average velocity is slope.
(1,69) and (1 + t,85(1 + t) - 16(1 + t)^2)
m = [ 85(1 + t) - 16(1 + t)^2 - 69 ] / [ 1 + t - 1 ]
m = [ 85(1 + t) - 16(1 + t)^2 - 69 ] / [ t ]
………………
t = 0.1
m = [ 85(1 + 0.1) - 16(1 + 0.1)^2 - 69 ] / [ 0.1 ]
m = [ 85(1.1) - 16(1.1)^2 - 69 ] / [ 0.1 ]
m = [ 85(1.1) - 16(1.21) - 69 ] / [ 0.1 ]
m = [ 93.5 - 19.36 - 69 ] / [ 0.1 ]
m = [ 5.14 ] / [ 0.1 ] = 51.4 ft/s
I'll let you finish the others … t = 0.01 and t = 0.001
………………………..
You'll guess what the velocity is when you find the other two. You should have have approaching values of 53ft/sec.
If you're in calculus, you can derivate the equation for the instantaneous velocity at 1 second. Then, the average velocity would be determined by:
Int[a,b] v(x) dx / (b - a)
s(x) [a,b] / (b - a)
(85t - 16t^2) [a,b] / (b - a)
(85b - 16b^2 - 85a + 16a) / (b - a)
(85b - 16b^2 - 85 + 16) / (b - 1)
(85b - 16b^2 - 69) / (b - 1)
Where b is the time after 1 second or 1 + t .. so:
(85(1 + t) - 16(1 + t)^2 - 69) / (1 + t - 1)
(85(1 + t) - 16(1 + t)^2 - 69) / (t)
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