Please help.
A particle moves in a stright line and its acceleration is given by a(t)=6t+4. its initial position is s(0)=9 and its position when t=1 is s(1)=6. find the velocity of the particle when t=2.
A. v(2)=14
B. v(2)=7
C. v(2)=4
D. v(2)=9
E. v(2)=0
a = 6 t+4
then
v = 3 t^2 + 4 t + c
then
s = t^3 + 2 t^2 + c t + k
now find the constants c and k from the given points
s(0) = 9
9 = 0^3 + 2 0^3 + c 0 + k
so
k = 9 and we so far have
s = t^3 + 2 t^2 + c t + 9
now
s(1) = 6
6 = 1^3 + 2 (1)^2 + c (1) + 9
6 = 1 + 2 + c + 9
c = -6
so in the end
s = t^3 + 2 t^2 - 6 t + 9
I guess you can put in t = 2
I get an answer of 13 but it's not one of the choices
To find the velocity of the particle when t = 2, we need to integrate the acceleration function with respect to time to obtain the velocity function. Since the acceleration function is given as a(t) = 6t + 4, integrating this function will give us the velocity function v(t).
Step 1: Integrate the acceleration function
Let's integrate a(t) = 6t + 4 with respect to time:
∫ (6t + 4) dt = 3t^2 + 4t + C
Step 2: Use the initial condition to find the constant of integration
Given that the initial position is s(0) = 9, we can use this information to determine the constant of integration (C) in the velocity function. Since velocity is the derivative of position, we can find v(0) using the position function:
s(t) = ∫ v(t) dt
Substituting s(0) = 9 into the position function:
9 = ∫ v(t) dt
Since the integral of v(t) with respect to t is the antiderivative of v(t), we can rewrite the equation as:
9 = ∫ (3t^2 + 4t + C) dt
9 = t^3 + 2t^2 + Ct + D
Step 3: Use the position at t = 1 to find C and D
Given that s(1) = 6, we can substitute t = 1 into the equation we obtained in step 2 and solve for C and D.
6 = 1^3 + 2(1^2) + C(1) + D
6 = 1 + 2 + C + D
9 = C + D (Equation 1)
Step 4: Find the velocity function
Now we have the constant of integration, and we can go back to the velocity function:
v(t) = 3t^2 + 4t + C
Using Equation 1, we can rewrite the velocity function as:
v(t) = 3t^2 + 4t + (9 - C)
Step 5: Find the velocity at t = 2
To find the velocity at t = 2, substitute t = 2 into the velocity function:
v(2) = 3(2)^2 + 4(2) + (9 - C)
v(2) = 12 + 8 + (9 - C)
v(2) = 29 - C
Now, we need to determine the value of C to find the velocity at t = 2.
Step 6: Use the position at t = 1 to solve for C
We know that s(1) = 6, so we can substitute t = 1 into the position function and solve for C.
6 = 1^3 + 2(1)^2 + C(1) + D
6 = 1 + 2 + C + D
9 = C + D (Equation 1)
Substituting D = 9 - C into the velocity function:
v(2) = 29 - C
v(2) = 29 - (9 - C)
v(2) = 29 - 9 + C
v(2) = 20 + C
Since C is unknown, we cannot determine the exact value of v(2) without more information.
Thus, the correct answer cannot be determined based on the given information.