Question

If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the following equation. S(t) = 76 + 128t − 16t2 (a) What is the average velocity (in ft/sec) in the first 4 seconds after it is thrown? ft/sec (b) What is the average velocity (in ft/sec) in the next 4 seconds?

Answer 1

a) Average velocity in first 4 seconds is 64 ft/s upward

b) Average velocity in second 4 seconds is 63.5 ft/s downward

Step-by-step explanation:

a) Given S(t) = 76 + 128t − 16t²

s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

Displacement in 4 seconds = 332 - 76 = 256 ft

Time = 4 - 0 = 4 s

`\texttt{Velocity = }(256)/(4)=64ft/s`

Average velocity in first 4 seconds is 64 ft/s upward

a) Given S(t) = 76 + 128t − 16t²

s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

Displacement in 4 seconds = 78 - 332 = -254 ft

Time = 4 - 0 = 4 s

`\texttt{Velocity = }(-254)/(4)=-63.5ft/s`

Average velocity in second 4 seconds is 63.5 ft/s downward