Question

If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet after t seconds is given by y=40t−16t2. (a) Find the average velocity for the time period beginning with t=2 and (1) lasting 0.5 seconds: -32 ft/s (2) lasting 0.1 seconds: -25.6 ft/s (3) lasting 0.05 seconds: -24.8 ft/s (4) lasting 0.01 seconds: -24.16 ft/s (b) Find the instantaneous velocity when t=2: ft/s

Answer 1

Answer:

(a) (1) -32ft/s

    (2) -25.6ft/s

    (3) -24.8ft/s

    (4) -24.16ft/s

(b) -24 ft/s

Explanation:

First of all, we need the position for every instant mentioned in order to calculate average velocities:

Y(2) = 2 ft

Y(2.5) = 0 ft

Y(2.1) = 13.44 ft

Y(2.05) = 14.76 ft

Y(2.01) = 15.7584 ft

Let's now calculate average velocities:

$V_{2-2.5}=\frac{Y(2.5)-Y(2)}{2.5-2}=-32ft/s$

$V_{2-2.1}=\frac{Y(2.1)-Y(2)}{2.1-2}=-25.6ft/s$

$V_{2-2.05}=\frac{Y(2.05)-Y(2)}{2.05-2}=-24.8ft/s$

$V_{2-2.01}=\frac{Y(2.01)-Y(2)}{2.01-2}=-24.16ft/s$

Now, for the instantaneous velocity, we derive the expression for y:

V(t) = 40 - 32t    For t=2s   V(2) = -24ft/s