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(Track and Field) The position (h) of a high jumper with respect to time is given by the function h(t)=-3.11t^(2)+9.33t. If h is in feet and t is in seconds, find the jumper's vertical speed in feet per second (fps), at t=1 seconds. In other words, find (dh)/(dt), the change in position over the change in time, which is the rate, or vertical speed of the jumper. Round to 1 d.p.

 Jun 6, 2022
 #1
avatar+124676 
+2

Take the derivative of the function

 

h' t =  -6.22t + 9.33

 

At 1 sec we have

 

h'(1)  =   -6.22 (1)  + 9.33  ≈   3.1  ft/s

 

 

cool cool cool

 Jun 6, 2022
 #2
avatar+490 
+1

Can you explain how you get the derivative? Please?

Elijah  Jun 6, 2022
 #3
avatar+124676 
+1

Derivative  of   - 3.11* t^2  =   2(-3.11) * t^(2-1)  =  - 6.22 * t  

 

Derivative of  9.33t^1  =   1*9.33* t^(1 - 1)  = 9.33t^0  = 9.33

 

General rule

 

Derivative of  a *t^n  =   n*a * t ^(n -1)

 

 

cool cool cool

CPhill  Jun 7, 2022

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