Rates of Change help, a simple problem.

• December 16th 2007, 03:07 PM
Jeavus
Rates of Change help, a simple problem.
A construction worker drops a bolt while working on a high rise building 320m above the ground. After t seconds, the bolt has fallen a distance of s metres, where s(t) = 320 - 5t^2, 0 < t < 8.

a. Find the avg. velocity during the first, third, and eighth seconds.
b. Find the avg. velocity for the interval 3 < t < 8.
c. Find the velocity at t = 2.

How do I do this problem?
• December 16th 2007, 06:13 PM
Soroban
Hello, Jeavus!

There is a typo in the problem . . .

Quote:

A construction worker drops a bolt while working on a 320-m building.
After $t$ seconds, the bolt has fallen a distance of $s$ metres,
where: . ${\color{blue}s(t) \:= \:320 - 5t^2},\quad0 < t < 8$ . . . . This is incorrect

a. Find the avg. velocity during the first, third, and eighth seconds.
b. Find the avg. velocity for the interval 3 < t < 8.
c. Find the velocity at t = 2.

The height of the bolt at time $t$ is: . $s(t) \:=\:320 - 5t^2$

a) At $t = 0:\;s(0) \:= \:320 - 5(0^2) \:=\:320\;m$
. . At $t = 1:\;s(1) \:=\:320 -5(1^2) \:=\:315\text{ m}$
In the first second, it has fallen: . $320 - 315 \:=\: 5\text{ m}$
. . Its average velocity is: . $\frac{5\text{ m}}{1\text{ sec}} \:=\:5\text{ m/sec}$

At $t = 2:\;s(2) \:=\:320-5(2^2) \:=\:300\text{ m}$
At $t = 3:\;s(3) \:=\:320 5(3^2) \:=\:275\text{ m}$
In the third second, it has fallen: . $300 - 275 \:=\:25\text{ m}$
. . Its average velocity is: . $\frac{25\text{ m}}{1\text{ sec}} \:=\:25\text{ m/sec}$

At $t = 7:\;s(7) \:=\:320 - 5(7^2) \:=\:75\text{ m}$
At $t = 8:\;s(8) \:=\:320 = 5(8^2) \:=\:0\text{ m}$
In the eighth second, it has fallen: . $75 - 0 \:=\: 75\text{ m}$
. . Its average velocity is: . $\frac{75\text{ m}}{1\text{ sec}} \:=\:75\text{ m/sec}$

$b) \;s(3) \:=\:275\;m,\quad s(8) \:= \:0$
The bolt fell: . $275 - 0 \:=\:275\text{ m}$ in 5 seconds.
. . Its average velocity is: . $\frac{275\text{ m}}{5\text{ sec}} \:=\:55\text{ m/sec}$

c) For velocity: . $v(t) \:=\:s'(t) \:=\:-10t$
. . At $t=2\!:\;v(2) \:=\:-10(2) \:=\:-20\text{ m/sec}$

(The minus indicates a downward velocity.)
.