## lexusbreon 2 years ago If a stone is thrown down at 100 ft/s from a height of 1,250 feet, its height after t seconds is given by s = 1,250 − 100t − 16t^2 Estimate its instantaneous velocity at time t = 2

1. swissgirl

We know that the equation for instantaneous velocity is $$f'(x)=\large \frac{f(x+h)-f(x)}{(x+h)-x}$$

2. swissgirl

Are you familiar with this equation?

3. lexusbreon

yes

4. swissgirl

so $$f(t)=1250-100t-16t^2$$ $$f(t+h)=1250-100(t+h)-16(t+h)^2$$ so $$f'(x)= \large \frac{(1250-100(t+h)-16(t+h)^2)-(1250-100t-16t^2)}{(x+h)-x}$$ $$f'(x)=\large \frac{1250-100t-100h-16t^2-32th-16h^2-1250+100t+16t^2}{h}$$ $$f'(x)=\large \frac{-100h-32th-16h^2}{h}$$ $$f'(x)= \large \frac{h(-100-32t-16h)}{h}$$ $$f'(x)=(-100-32t-16h)$$

5. swissgirl

Ok Are you familiar with limits? Like what does f'(t) equal as h approaches 0?

6. swissgirl

Did ya follow so far?

7. lexusbreon

Yes I follow so far

8. swissgirl

Ok and keep in mind by accident I used X's in stead of T's so just ignore that

9. swissgirl

|dw:1373424544924:dw|

10. swissgirl

ok so the difference btwn x and x+h is just h

11. swissgirl

Basically we have 2 points; x and x+h and we want them to beee sooooo close together So we want h to be realllllly small So we want h to be like nearly 0 Like the number right next to 0 So there is this method called limits and we limit the equation meaning we want to see what the equation will equal as h APPROACHES 0 $$\large f'(t)_{\lim h \to 0} -100-32t-16h$$ Now there is only one term that has h in it which is -16h Now as h approaches 0 then -16h approaches 0 since 16*.00000000001=.00000000016 So its basically 0 so we consider it as if it is 0 and just trash that whole term $$\large f'(t)_{\lim h \to 0} -100-32t-16h=-100-32t$$ Now we need to find the Instantaneous velocity when t=2 So we plug into our equation t=2 \(f'(2)=-100-32(2)=-164

12. swissgirl

So at t=2 the stone is travelling -164 ft/s downwards

13. swissgirl

I hope you followed :)

14. lexusbreon

Thank You so much!!