If a ball is thrown into the air with a velocity of 46 ft/s, its height in feet t seconds later is given by y = 46t − 16t2.

Question

(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

Answer 1

Average velocity would be [ ∫v(t) dt ] / Δt. You can find v = 46 -32t from derivative, but when you integrate, you're right back to the expression for y(t) = 46t - 16t². So ave v = displacement/time interval = [y_b - y_a] / [t_b - t_a]. Initial position y_a = y(2) = 28 ft.

i) t_b = 2.5, y(2.5) = 15 ft, ave v = (-13) / (0.5) = -26 ft/s.

ii) t_b = 2.1, y(2.1) = 26.04 ft, ave v = (-1.96) / (0.1) = -19.6 ft/s.

iii) t_b = 2.05, y(2.05) = 27.06 ft, ave v = (-0.94) / (0.05) = -18.8 ft/s.