Question

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 46 ft/s. Its height in feet after t seconds is given by y=46t−19t2. A. Find the average velocity for the time period beginning when t=3 and lasting
.01 s:
.005 s:
.002 s:
.001 s:

Answer 1

Answer:

The average velocities are:

for 0.01.s

• $<v> = - 68.19 \frac{ft}{s} \hat{j}$

for 0.005 s

• $<v> = - 68.2 \frac{ft}{s} \hat{j}$

for 0.002 s

• $<v> = - 68.05 \frac{ft}{s} \hat{j}$

for 0.001 s

• $<v> = - 68.02 \frac{ft}{s} \hat{j}$

Explanation:

The average velocity is given by

$<v> = \frac{displacement}{time}$

So, we just need to find the position at t = 3 s and then, after every period of time.

Position at t = 3 s

Knowing that

$y(t) = 46 \frac{ft}{s} t - 19 \frac{ft}{s^2} \ t^2$

at t = 3 s we have

$y(3 \ s) = 46 \frac{ft}{s} * 3 s - 19 \frac{ft}{s^2} * (3 \ s)^2$

$y(3 \ s) = 138 ft - 19\frac{ft}{s^2} * 9 \ s^2$

$y(3 \ s) = -33 ft$

After 0.01 s

After 0.01 s the position will be

$y(3.01 s) = 46 \frac{ft}{s} 3.01 s - 19 \frac{ft}{s^2} \ (3.01 s)^2$

$y(3.01 s) = -33.6819 ft$

So, the average velocity will be

$<v> = \frac{-33.6819 ft \hat{j}- (-33 ft) \hat{j}}{0.01 s}$

$<v> = \frac{-0.6819 ft \hat{j}}{0.01 s}$

$<v> = - 68.19 \frac{ft}{s} \hat{j}$

The minus sign is there cause the velocity is pointing downward.

After 0.005 s

After 0.005 s the position will be

$y(3.005 s) = 46 \frac{ft}{s} 3.005 s - 19 \frac{ft}{s^2} \ (3.005 s)^2$

$y(3.005 s) = -33.3405 ft$

So, the average velocity will be

$<v> = \frac{-33.3405 ft \hat{j} - (-33 ft) \hat{j} }{0.005 s}$

$<v> = \frac{-0.3405 ft }{0.005 s} \hat{j}$

$<v> = - 68.2 \frac{ft}{s} \hat{j}$

After 0.002 s

After 0.002 s the position will be

$y(3.002 s) = 46 \frac{ft}{s} 3.002 s - 19 \frac{ft}{s^2} \ (3.002 s)^2$

$y(3.002 s) = -33.1361 ft$

So, the average velocity will be

$<v> = \frac{-33.1361  ft \hat{j} - (-33 ft) \hat{j} }{0.002 s}$

$<v> = \frac{-0.1361  ft \hat{j} }{0.002 s}$

$<v> = - 68.05 \frac{ft}{s} \hat{j}$

After 0.001 s

After 0.001 s the position will be

$y(3.001 s) = 46 \frac{ft}{s} 3.001 s - 19 \frac{ft}{s^2} \ (3.001 s)^2$

$y(3.001 s) = -33.06802 ft$

So, the average velocity will be

$<v> = \frac{-33.06802  ft  \hat{j} - (-33) ft \hat{j} }{0.001 s}$

$<v> = \frac{-0.06802  ft \hat{j} }{0.001 s}$

$<v> = - 68.02 \frac{ft}{s} \hat{j}$