Question

The deck of a bridge is suspended 275 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 = 275 - 16t^2.
(a) Find the average velocity of the pebble for the time period beginning when t − 4 and lasting (i) 0.1 seconds (ii) 0.05 seconds (iii) 0.01 seconds (b) Estimate the instantaneous velocity of the pebble after 4 seconds.

Answer 1

Answer:

Our equation for the height is:

y(t) = 275 - 16*t^2.

a) To find the average velocity between two times, t1 and t2, (where t2 > t1) the equation is:

$AV = \frac{y(t2) - y(t1)}{t2 - t1}$

Then:

i) t1 = 4s, t2 = 4s + 0.1s = 4.1s

The average velocity is:

$AV = \frac{(275 - 16*4.1^2) - (275 - 16*4^2)}{4.1 - 4} = \frac{16(4^2 - 4.1^2)}{0.1} = -129.6$

And the units will be ft/s, so the average speed is:

-129.6 ft/s

The minus sign is because te pebble is falling down.

ii)  t1 = 4s, t2 = 4s + 0.05s = 4.05s

The average velocity is:

$AV = \frac{(275 - 16*4.05^2) - (275 - 16*4^2)}{4.05 - 4} = \frac{16(4^2 - 4.05^2)}{0.05} = -128.8$

So the average speed is -128.9 ft/s

iii)  t1 = 4s, t2 = 4s + 0.01s = 4.01s

The average speed is:

$AV = \frac{(275 - 16*4.01^2) - (275 - 16*4^2)}{4.01 - 4} = \frac{16(4^2 - 4.01^2)}{0.01} = -128.16$

The average speed is -128.16 ft/s.

b) The instantaneous velocity of the pebble after 4 seconds can be obtained by looking at the velocity equation, that is the derivative of the height equation.

v(t) = dy(t)/dt.

v(t) = -2*16*t + 0

Then the velocity at t = 4s is:

v(4s) = -32*4 = -128

The instantaneous velocity at t = 4s is -128 ft/s.