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Math  /  Calculus

QuestionA particle is moving with the given data. Find the position of the part (b) a(t)=2t+1s(0)=3,v(0)=−La(t)=2 t+1 s(0)=3, v(0)=-L

Studdy Solution

STEP 1

1. The acceleration function is given by a(t)=2t+1 a(t) = 2t + 1 .
2. The initial position of the particle is s(0)=3 s(0) = 3 .
3. The initial velocity of the particle is v(0)=−2 v(0) = -2 .

STEP 2

1. Integrate the acceleration function to find the velocity function.
2. Use the initial velocity to determine the constant of integration for the velocity function.
3. Integrate the velocity function to find the position function.
4. Use the initial position to determine the constant of integration for the position function.
5. Write the final expression for the position function.

STEP 3

Integrate the acceleration function a(t)=2t+1 a(t) = 2t + 1 to find the velocity function v(t) v(t) :
v(t)=∫(2t+1) dt v(t) = \int (2t + 1) \, dt

STEP 4

Perform the integration:
v(t)=∫(2t+1) dt=t2+t+C1 v(t) = \int (2t + 1) \, dt = t^2 + t + C_1

STEP 5

Use the initial velocity v(0)=−2 v(0) = -2 to find C1 C_1 :
v(0)=02+0+C1=−2 v(0) = 0^2 + 0 + C_1 = -2 C1=−2 C_1 = -2

STEP 6

Substitute C1 C_1 back into the velocity function:
v(t)=t2+t−2 v(t) = t^2 + t - 2
Integrate the velocity function to find the position function s(t) s(t) :
s(t)=∫(t2+t−2) dt s(t) = \int (t^2 + t - 2) \, dt

STEP 7

Perform the integration:
s(t)=∫(t2+t−2) dt=t33+t22−2t+C2 s(t) = \int (t^2 + t - 2) \, dt = \frac{t^3}{3} + \frac{t^2}{2} - 2t + C_2

STEP 8

Use the initial position s(0)=3 s(0) = 3 to find C2 C_2 :
s(0)=033+022−2(0)+C2=3 s(0) = \frac{0^3}{3} + \frac{0^2}{2} - 2(0) + C_2 = 3 C2=3 C_2 = 3

STEP 9

Substitute C2 C_2 back into the position function:
s(t)=t33+t22−2t+3 s(t) = \frac{t^3}{3} + \frac{t^2}{2} - 2t + 3
The position function of the particle is:
s(t)=t33+t22−2t+3 s(t) = \frac{t^3}{3} + \frac{t^2}{2} - 2t + 3

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