The height above ground,h, of an arrow t seconds after being shot straight in the air can be modeled by the formula h(t)=1.5+40t-4.9t^2, t= seconds h=meters. What is average rate of change of the height,in meters per second,of the first three second after being shot?

Question

The height above ground,h, of an arrow t seconds after being shot straight in the air can be modeled by the formula  h(t)=1.5+40t-4.9t^2, t= seconds h=meters. What is average rate of change of the height,in meters per second,of the first three second after being shot?

amistre64 · Answer

$$\frac{h(3)-h(0)}{3-0}$$

amistre64 · Answer

in general, the slope of the line between (0,h(0))  and (3,h(3))

anonymous · Answer

what do you do with the equation?

phi · Answer

you are given 
 h(t)=1.5+40t-4.9t^2

to find h(3),   erase the t in the equation, and put in 3 instead.  then do the arithmetic
for h(0), do the same thing, except use 0.

example:  h(0)=  1.5 + 40*0 - 4.9 * 0^2
if you simplify  you get h(0)= 1.5+ 0 - 0 = 1.5
so h(0)= 1.5

can you find h(3) ?

anonymous · Answer

yes thank you