Question

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t second is given by y=70t−16t2. Find the average velocity for the time period beginning when t = 1and lasting:_____

Answer 1

the average velocity for the time period beginning when t=1 and lasting Δt is 70 - 32(t+Δt) - 38Δt feet per second.

Step-by-step explanation:

average velocity = (change in displacement) / (change in time)

In this case, the displacement is the change in height of the ball during the time period, and the change in time is Δt. So we need to find the height of the ball at the end of the time period (t=1+Δt) and subtract the height of the ball at the beginning of the time period (t=1), and then divide by the length of the time period (Δt).

Therefore, we have:

Height at end of time period: y = 70(t+Δt) - 16(t+Δt)^2

Height at beginning of time period: y = 70(1) - 16(1)^2

Subtracting the two heights and dividing by the length of the time period, we get:

average velocity = (70(t+Δt) - 16(t+Δt)^2 - 54) / Δt

Simplifying and factoring, we get:

average velocity = 70 - 32(t+Δt) - 38Δt

Therefore, the average velocity for the time period beginning when t=1 and lasting Δt is 70 - 32(t+Δt) - 38Δt feet per second.