Question

If a ball is thrown into the air with a velocity of 38 ft/s, its height in feet t seconds later is given by y = 38t − 16t2. (a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following.
(i) 0.5 seconds ft/s

(ii) 0.1 seconds ft/s

(iii) 0.05 seconds ft/s

(iv) 0.01 seconds ft/s

(b) Estimate the instantaneous velocity when t = 2. ft/s

Answer 1

Answer:

i.-34ft/s

ii.-27.6ft/s

iii.-26.8ft/s

iv-26.7ft/s

B.-26ft/s

Step-by-step explanation:

$average velocity = \frac{y_{2}-y_{1}}{t_{2}-t_{1}}$

For all cases the value of the initial time $t_{1} =2 seconds$

now we also use this time ti determine the value of the initial height in all cases

$y_{1}=38(2)-16(2)^{2} \\y_{1}= 12ft$

now when the time is increase by 0.5 seconds  the new height becomes

$y_{2}=38(2.5)-16(2.5)^{2} \\y_{2}=-5fts$

the velocity becomes

$[tex]Average velocity = \frac{-5-12}{2.5-2}\\Average velocity = -34ft/s$

ii. when the time is increase by 0.1 seconds

the new height becomes

$y_{2}=38(2.1)-16(2.1)^{2} \\y_{2}= 10.66ft$

$Average velocity = \frac{10.66-12}{2.05-2}\\Average velocity = -27.6ft/s$

iii. when the time is increase by 0.05 seconds

the new height becomes

$y_{2}=38(2.05)-16(2.05)^{2} \\y_{2}= 9.24ft$

$Average velocity = \frac{9.24-12}{2.05-2}\\Average velocity = -26.8ft/s$

iv.  when the time is increase by 0.1 seconds

the new height becomes

$y_{2}=38(2.01)-16(2.01)^{2} \\y_{2}= 11.73ft$

$Average velocity = \frac{11.73-12}{2.01-2}\\Average velocity = -26.7ft/s$.

B. For us to determine the instantaneous velocity expression, we differentiate the expressing for the height

$V_{inst}=38-32t$

we now substitute t=2, we arrive at

$V_{inst}=-26ft/s$