Question

The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of the pebble above the water surface after t seconds is given by y = 235 − 16t^2.
Required:
a. Find the average velocity (in ft/s) of the pebble for the time period beginning when t = 2 and lasting the following amount of time.

1. 0.1 sec:________
2. 0.05 sec:_______
3. 0.01 sec:_______

b. Estimate the instantaneous velocity (in ft/s) of the pebble after 3 seconds. ft/s.

Answer 1

Answer:

(a) 1, average velocity = -65.6 m/s

   2, average velocity = -64.8 m/s

   3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given  by;

$y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}$

(1)

$y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s$

(2)

$y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s$

(3)

$y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s$

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s