Find the distance, in feet, a particle travels in its first 2 seconds of travel, if it moves according to the velocity equation v(t)= 6t2 − 18t + 12 (in feet/sec).
4
5<-my answer
6
−1
The answer is 6 i just got i right
its not 4
I marked 5 and got it wrong.... Based on the Chegg answers, I have a feeling 4 is right.
To find the distance that a particle travels in its first 2 seconds of travel, we need to integrate the velocity equation over the interval [0, 2] since the velocity gives us the rate of change of displacement.
First, let's find the antiderivative of the velocity equation, v(t), to get the displacement equation, s(t):
s(t) = ∫(6t² - 18t + 12) dt
Taking the integral of each term individually, we get:
s(t) = 2t³ - 9t² + 12t + C
Now, to find the constant of integration, C, we need to use the initial condition of the particle's position. However, since the position of the particle is not given, we can ignore the constant of integration for now and focus on finding the displacement over the interval [0, 2].
To find the displacement, s(2) - s(0), we substitute t = 2 into our displacement equation and subtract the result when t = 0:
s(2) - s(0) = (2(2)³ - 9(2)² + 12(2)) - (2(0)³ - 9(0)² + 12(0))
s(2) - s(0) = (16 - 36 + 24) - (0 - 0 + 0)
s(2) - s(0) = 4
Therefore, the particle travels a distance of 4 feet in its first 2 seconds of travel.
So, none of the answer choices provided - 4, 5, 6, -1 - are correct.