The formula d = 1.1 t 2 + t + 1 expresses a car's distance (in feet to the north of an intersection, d, in terms of the number of seconds t since the car started to move. As the time t since the car started to move increases from t = 3 to t = 5 seconds, what constant speed must a truck travel to cover the same distance as the car over this 2-second interval?

9.8 ft/s

Explanation:

The formula for the distance (d, in ft) traveled by a cer to the north of an intersection is:

d = 1.1 t² + t + 1

where,

t is the time in seconds

The distance traveled in the first 3 seconds is:

d (3) = 1.1 * (3) ² + 3 + 1 = 13.9 ft

The distance traveled in the first 5 seconds is:

d (5) = 1.1 * (5) ² + 5 + 1 = 33.5 ft

The distance traveled in the 3-5 s interval is:

d (5) - d (3) = 33.5 ft - 13.9 ft = 19.6 ft

A truck should cover the same distance (19.6 ft) in the same time (2s). It must have a constant speed of:

v = d/t = 19.6 ft/2s = 9.8 ft/s

d (t) = 1.1t² + t + 1

The constant speed required to cover the same distance between t = 3 to t = 5 is the same as the average speed over that same time interval. It is given by:

v = Δx/Δt

v = average speed, Δx = change in distance, Δt = elapsed time

Given values:

Δx = d (5) - d (3) = 19.6ft

Δt = 5s - 3s = 2s

Plug in and solve for v:

v = 19.6/2

v = 9.8ft/s