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Answer:
219 ft
Explanation:
Here we can define the value t = 0s as the moment when the car starts decelerating.
At this point, the acceleration of the car is given by the equation:
A(t) = -10 ft/s^2
Where the negative sign is because the car is decelerating.
To get the velocity equation of the car, we integrate over time, to get:
V(t) = (-10 ft/s^2)*t + V0
Where V0 is the initial velocity of the car, we know that this is 88 ft/s
Then the velocity equation is:
V(t) = (-10 ft/s^2)*t + 88ft/s
To get the position equation we need to integrate again, this time we get:
P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0
Where P0 is the initial position of the car, we do not know this, but it does not matter for now.
We want to find the total distance that the car traveled in a 3 seconds interval.
This will be equal to the difference in the position at t = 3s and the position at t = 0s
distance = P(3s) - P(0s)
= ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)
= ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)
= (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft
The car advanced a distance of 219 ft in the 3 seconds interval.