# Velocity rate question

1. Nov 11, 2012

### fifaking7

1. The problem statement, all variables and given/known data
s(t)= Vit + 1/2 a t^2 + So

a) find the average velocity on the interval [1,3]
b) find the instantaneous velocity when t= 3s.
c) how long does it take the coin to hit the ground?
d) find the velocity when the coin hits the ground

2. Relevant equations
a= -32ft/(s^2)

3. The attempt at a solution
S(t)=Vi(0)t + 1/2 (-32ft/s^2)t^2 +So
S(t)= 0 -16t^2 + 800
v(t) = -32t

a) V(3)-V(1)/(3-1)
V(3) = -32(3)=-96
V(1) = -32(1)= -32

-96-(-32)/2 = -64ft/sec is my final answer

b) v(3)= -32(3)

C)coin hits ground at s(t)= 0 S(t)= -16t^2 + 800
0= =16t^2 +800
-800= -16t^2
sqr50= t^2
7.071 seconds is when the coin hits the ground
D) V(t) =-32t
V(7.071)= -32(7.071)
v= -226.272 ft/sec

2. Nov 11, 2012

### Abhinav R

1) Yes the average velocity can be calculated by differentiating s(t)
∴ s(t$^{'}$) = -32t
The average velocity turns out to be (-96 + 32)/2 = -32

2) The instantaneous velocity when t=3 s is given by
Lim Δt→0 Δs/Δt = ds/dt
So -32t = -32 × 3 = -96

3) s(t$^{'}$) = -32t
∴ s(t) = $\int$ -32t dt
= -16t$^{2}$ + C

4) Velocity is zero as when t=0 ; -32t = -32 × 0 = 0

3. Nov 11, 2012

### haruspex

There seem to be other facts which you introduce later. From those I infer a coin is dropped from rest at time 0 and a height of 800ft.
No, that will give you the average of two specific speeds. That is not the same as the average speed over an interval. Average speed = change in distance / change in time.
Abhinav R's method and answer are wrong too.

4. Nov 12, 2012

### fifaking7

How do I do it then if the methods are wrong?

5. Nov 12, 2012

### SammyS

Staff Emeritus
Please state the complete problem as it was given to you.

6. Nov 12, 2012

### haruspex

As I said, Average speed = change in distance / change in time. Where was it at t=1? Where at t=3? How far did it travel between t=1 and t=3?