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 equation editorEquation Editor ft/s (2) lasting 0.1 seconds: -25.6 equation editorEquation Editor ft/s (3) lasting 0.05 seconds: -24.8 equation editorEquation Editor ft/s (4) lasting 0.01 seconds: -24.16 equation editorEquation Editor ft/s (b) Find the instantaneous velocity when t=2

Answer 1

Answer:

a) $v_{1} =-32ft/s\\v_{2} =-25.6ft/s\\v_{3} =-24.8ft/s\\v_{4} =-25ft/s$

b) $v=-24ft/s$

Explanation:

The equation of the ball's motion is given by

$y=40t-16t^{2}$

a) To find the average velocity we need to use the following formula:

$v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }$

Since we know t=2,t=2.5,t=2.1,t=2.05,t=2.01 we need to find its position at these times.

$y_{t=2}=40(2)-16(2)^{2}=16ft$

$y_{t=2.5}=40(2.5)-16(2.5)^{2}=0ft$

$v_{1} =\frac{(0-16)ft}{(2.5-2)s}=-32ft/s$

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$y_{t=2.1}=40(2.1)-16(2.1)^2=13.44ft$

$v_{2} =\frac{(13.44-16)ft}{(2.1-2)s}=-25.6ft/s$

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$y_{t=2.05}=40(2.05)-16(2.05)^2=14.76ft$

$v_{3} =\frac{(14.76-16)ft}{(2.05-2)s}=-24.8ft/s$

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$y_{t=2.01}=40(2.01)-16(2.01)^2=15.75ft$

$v_{4} =\frac{(15.75-16)ft}{(2.01-2)s}=-25ft/s$

b) To find the instantaneous velocity when t=2 we need to derivate the equation of position

$v=\frac{dy}{dt}=40-32t$

$v=40-32(2)=-24ft/s$