v = ft/s
dv/dt is thus ft/s^2 (acceleration)
Since gravity is constant, v' is constant.
The velocity of a ball that has been tossed vertically in the air is given by v(t) = 16 − 32t, where v is measured in feet per second, and t is measured in seconds. The ball is in the air from t = 0 until t = 2.
Determine the value of v'(1).
What are the units on the value of v'(1)?
Thanks!
2 answers
since v'(t) = a(t), which is the acceleration
a(t) = -32
since a(t) is a constant, we get the acceleration due to gravity at -32 ft/s^2
and v'(1) = a(1) = -32 ft/s^2
a(t) = -32
since a(t) is a constant, we get the acceleration due to gravity at -32 ft/s^2
and v'(1) = a(1) = -32 ft/s^2
1 answer