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).
A. 4
B. 5
C. 6
D. -1
Please helpp, I'm struggling!
Thank you in advance for your help!
good guess.
To find the distance traveled in the first 2 seconds, we need to calculate the definite integral of the velocity function from 0 to 2.
The definite integral of the velocity function represents the area under the curve of the function, which gives us the displacement or distance traveled.
To find the definite integral of the velocity function, we can use the power rule for integration.
The power rule states that the integral of x^n with respect to x is (x^(n+1))/(n+1), where n is any real number except -1.
Applying the power rule to each term of the velocity function, we get:
∫(6t^2 - 18t + 12) dt = (6/3) t^3 - (18/2) t^2 + (12/1) t + C
Simplifying this expression, we have:
2t^2 - 9t^2 + 12t + C
Evaluating the definite integral from 0 to 2, we substitute the upper and lower limits:
[2(2)^3 - 9(2)^2 + 12(2)] - [2(0)^3 - 9(0)^2 + 12(0)] = 16 - 36 + 24 - 0 = 4
Therefore, the particle travels a distance of 4 feet in its first 2 seconds of travel. Hence, the correct answer is option A.
v(t) = ds/dt
so, the distance s is
s(t) = ∫ v(t) dt = 2t^3 - 9t^2 + 12t
So, what do you think?
So I guess now that we have the integral, I substitute the t=2 into the equation, right?
I got and selected 4 after substituting t=2, but when I selected that one I got marked wrong...
I get 4 as well.
Better check for typos. Also, answer keys have been known to be wrong ...