1. A body is moving along a straight line with a velocity which varies according to the equation v=9t^2+2t, where v is in feet per second and t is in seconds. Find the expression for the distance as a function of time.
A. 3t^3+t^2+C
B. 3t^3+2t^2+C
C. 18t+C
D. 20t+C
My answer is B.
recall that velocity is the change in position over change in time, or:
v = dx/dt
where x=distance, and t=time
thus equating this to the given function,
dx/dt = 9t^2 + 2t
dx = (9t^2 + 2t)dt
integrating,
x - xo = 3t^3 + t^2 + C
if we assume that the initial displacement is zero or xo = 0, this becomes:
x = 3t^3 + t^2 + C
your answer is correct~! :D
I don't know how to do this
To find the expression for the distance as a function of time, we need to integrate the velocity function.
Given: v = 9t^2 + 2t
To integrate the velocity function, we can use the power rule of integration:
∫v dt = ∫(9t^2 + 2t) dt
Applying the power rule, we get:
∫v dt = (9/3)t^3 + (2/2)t^2 + C
Simplifying the expression further:
∫v dt = 3t^3 + t^2 + C
Therefore, the expression for the distance as a function of time is:
Distance = 3t^3 + t^2 + C
So, the correct answer is A, not B.
To find the expression for the distance as a function of time, we need to integrate the velocity function with respect to time. The velocity function is given as v = 9t^2 + 2t.
To integrate, we can use the power rule for integration. The power rule states that the integral of x^n with respect to x is (1/n+1) * x^(n+1).
Applying the power rule to our velocity function, we get:
∫ (9t^2 + 2t) dt = (9/3) * t^3 + (2/2) * t^2 + C
Simplifying:
3t^3 + t^2 + C
Therefore, the expression for the distance as a function of time is 3t^3 + t^2 + C.
So, the correct answer is A, not B.