The velocity of a particle at time t is v(t)=t^3. When t=0, the particle is at position x=3. Determine the location of the particle at time t=2.
A)x=4
B)x=7
C)x=8
D)x=16
E)x=19
v(t)=t^3
s(t) = (4/3)t^4 + c
s(0) =3
3 = 0 + c
s(t) = (4/3)t^3 + 3
s(2) = (4/3)(8)+3
= 41/3
that should have been:
s(t) = (1/4)t^4 + c
...
s(t) = (1/4)t^4 + 3
s(2) = (1/4)(16) + 3
= 7
please ignore the gibberish I had above
To determine the location of the particle at time t=2, we need to integrate the velocity function v(t) from t=0 to t=2.
The position function x(t) is obtained by integrating the velocity function v(t).
∫v(t) dt = ∫t^3 dt
To evaluate this integral, we use the power rule for integration, which states that if the derivative of g(t) is t^n, then the integral of t^n dt is (1/(n+1)) t^(n+1).
Applying the power rule, we have:
∫t^3 dt = (1/(3+1)) t^(3+1) = (1/4) t^4
Now, we can evaluate the definite integral from t=0 to t=2:
∫[0 to 2] t^3 dt = [(1/4) t^4] [0 to 2] = (1/4) (2^4) - (1/4) (0^4)
= (1/4) (16) - (1/4) (0)
= 4 - 0
= 4
Therefore, the location of the particle at time t=2 is x=4.
Hence, the correct answer is option A) x=4.
To determine the location of the particle at time t=2, you need to find the position function x(t) by integrating the velocity function v(t).
Given that the velocity function is v(t) = t^3, integrating this function will give us the position function x(t).
To integrate t^3, you add 1 to the power and divide by the new power. So, integrating t^3 gives us (1/4)*t^4.
Now we can find the constant of integration by using the initial condition given: when t=0, x=3. So plugging these values into the position function, we have:
(1/4)*0^4 + C = 3
0 + C = 3
C = 3
Therefore, the position function x(t) is (1/4)*t^4 + 3.
To find the location of the particle at time t=2, we can substitute t=2 into the position function:
x(2) = (1/4)*2^4 + 3
x(2) = (1/4)*16 + 3
x(2) = 4 + 3
x(2) = 7
Therefore, the location of the particle at time t=2 is x=7.
So the correct answer is B) x=7.