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

v_av= [y(t + Δt) - y(t)]/Δt

i) t = 2s and Δt = 0.5 s, then v_av = [(46·2.5 - 16·2.5²) - (46·2 - 16·2²)]/0.5 = - 26 ft/s

ii) t = 2s and Δt = 0.1 s, then v_av = [(46·2.1 - 16·2.1²) - (46·2 - 16·2²)]/0.1 = - 19.6 ft/s

iii) t = 2s and Δt = 0.05 s, then v_av = [(46·2.05 - 16·2.05²) - (46·2 - 16·2²)]/0.05 = - 18.8 ft/s