A .
correct
B.
for the average velocity no Calculus is needed
P(1) = -16(1) + 1454 = 1438
P(3) = -16(9) + 1454 = 1310
average velocity = (1310 - 1438)/(3-1) = -64 ft/s
C. now you want to set
-32t = -64
t = 2
E. correct
F. I disagree.
you want P(t) = 0
-16t^2 + 1454 = 0
t^2 = 90.875
t = 9.532 , you had that, but gave it the wrong units.
You want the velocity when t = 9.532
velocity = -32(9.532) = -305.05 ft/s
Can you please check my answers and help me fix the one I don't or or did wron .
The height in feet of a free falling object t seconds after release is s(t)=-16t^2+ v_0t+s_0, where s_0 is the height(in feet) at which the object is realsed, and v_0 is the initial velocity (in feet per second). Suppose the coin is dropped from a height of 1454 feet.
A. Determine the position and velocity functions
P(t)=-16t^2+1454
P'(t)=-32t
B.Determine the average velocity on the interval [1,3]
How would you go about this I tried to find the slope of the secant line and subtracted s(3)-s(2)/3-2 and got -64ft/sec
C.Find the instaneous velocity when t=1 and t=3
T=3 -96 ft/sec
T=1 -32 ft/sec
D.At what time is the instaneous velocity of the coin equal to the average velocity of the coin found in part B?
Need help here
E. What is the name of the theorwm that says there must be at least one solution to part D?
Mean Value thm
F. Find the velocity of the coin just before it hits the ground.
T=9.53284 ft/sec
1 answer
1 answer