The formula d = 1.1 t 2 + t + 2 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 = 2 to t = 6 seconds, what constant speed must a truck travel to cover the same distance as the car over this 4-second interval? feet per second as the time t since the car started to move increases from t = 7 to t = 7.1 seconds, what constant speed must a truck travel to cover the same distance as the car over this 0.1-second interval?

The speed of the truck must be 9.8 ft/s.

The speed of the truck must be 17 ft/s

Explanation:

First, let´s calculate the distance of the car relative to the intersection at time t = 2 s and t = 6 s:

d(t) = 1.1 t² + t +2

d(2) = 1.1 (2)² + 2 + 2 = 8.40 ft

d(6) = 1.1(6)² + 6 + 2 = 47. 6 ft

The car traveled (47.6 ft - 8.40 ft) 39.2 ft in 4 seconds.

Then, the speed of the truck must be (39.2 ft / 4 s) 9.8 ft/s to cover the same distance as the car in 4 seconds.

Now, let´s find the distance of the car to the intersection at time t = 7 and t = 7.1.

d(7) = 1.1(7)² + 7 + 2 = 62.9 ft

d(7.1) = 1.1(7.1)² + 7.1 + 2 = 64.6 ft

The distance traveled by the car in 0.1 s is ( 64.6 ft - 62.9 ft) 1.7 ft

Then the truck must travel with a velocity of (1.7 ft /0.1s) 17 ft/s to cover that distance in 0.1 s.  

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