rate of change, given h(t) = 2 + 20t - 5t^2, after 3 seconds

[adam2016]

Hi guys,

so I'm asked to find the rate of when a ball is rising after 3 seconds given the equation h(t) = 2 + 20t - 5t^2

so I attempted the question first I selected an x value = 0 and subbed it into the above equation to get the y value( 2 + 20(0) - 5(0)^2) the answer came out to be 2 (obviously)

so I then set x = 1 and got the corresponding y value which I got 17 ( 2 + 20(1) - 5(1)^2)

so I subbed this into the formula y2 - y1 / x2 - x1 with 17 being y2,x2 being 1,y1 2,x1 0

so the average rate of change I got was 15, I thought to myself this is pretty high so I then tied to sub in 2 and 3 to see if I got the same rate of change by subbing in 2 and 3 into the formula as x I got y1 == 2, y2 = 53

with the rate of change formula being 53 - 22/3 - 2 = 31, so since I got two different rates of change this clearly can't be the right answer,

I tried just just differentiated the equation and subbed in 3 to get -10 and this clearly can't be the right answer either,

where am I going wrong?

thanks

 

[mmm4444bot]

rate of when a ball is rising …

… I tried just just differentiated the equation and subbed in 3 to get -10 and this clearly can't be the right answer …

Hello there!

As you know how to differentiate, you ought to have done that right away.

You did not provide a complete exercise statement (please familiarize yourself with our guidelines). Is that really what they told you?

Assuming the units on h(t) are meters, then -10 m/s is the correct answer.

Is it the negative sign that concerns you? Velocity is a vector quantity, so the negative sign means 10 m/s in the downward direction. :cool: