average velocity over the time interval

• Sep 1st 2009, 06:07 PM
wnsghkghk
average velocity over the time interval
A stone is tossed in the air from ground level with an initial velocity of 15m/s. Its height at time t is h(t) = 15t - 4.9t^2 m.

Compute the stone's average velocity over the time interval [1, 3.5].Average velocity https://webwork.math.lsu.edu/webwork...bf9402a711.png ?

I tried to solve it, but when time is '3.5' v became minus...
• Sep 1st 2009, 06:18 PM
skeeter
Quote:

Originally Posted by wnsghkghk
A stone is tossed in the air from ground level with an initial velocity of 15m/s. Its height at time t is h(t) = 15t - 4.9t^2 m.

Compute the stone's average velocity over the time interval [1, 3.5].Average velocity https://webwork.math.lsu.edu/webwork...bf9402a711.png ?

I tried to solve it, but when time is '3.5' t became minus...

$\bar{v} = \frac{h(3.5) - h(1)}{3.5 - 1}
$

$\bar{v} = \frac{h(3.5) - h(1)}{3.5 - 1} = -7.05$ m/s
• Sep 1st 2009, 06:24 PM
HallsofIvy
Quote:

Originally Posted by wnsghkghk
A stone is tossed in the air from ground level with an initial velocity of 15m/s. Its height at time t is h(t) = 15t - 4.9t^2 m.

Compute the stone's average velocity over the time interval [1, 3.5].Average velocity https://webwork.math.lsu.edu/webwork...bf9402a711.png ?

I tried to solve it, but when time is '3.5' t became minus...

"t" is 3.5 which is NOT negative. I presume you mean that v is negative- and there is nothing wrong with that.
• Sep 1st 2009, 08:15 PM
wnsghkghk
ughhhhhhhhh I put just 7.05 not -7.05