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Sia

mzz's Question
Math
Posted 8 months ago

A particle, initially at rest, moves along the xx-axis so that its acceleration at any time t0t \geq 0 is given by a(t)=12t24a(t)=12 t^{2}-4. The position of the particle when t=1t=1 is x(1)=3\boldsymbol{x}(1)=3.
(a) Find the values of tt for which the particle is at rest.
(b) Write an expression for the position x(l)\boldsymbol{x}(\boldsymbol{l}) of the particle at any time t0t \geq 0.
(c) Find the total distance traveled by the particle from t=0t=0 to t=2t=2.
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Answer from Sia
Posted 8 months ago
Solution by Steps
step 1
The particle is at rest when its velocity v(t)=0v(t) = 0. To find v(t)v(t), we integrate the acceleration function a(t)=12t24a(t) = 12t^2 - 4
step 2
Integrate a(t)a(t): (12t24)dt=4t34t+C\int (12t^2 - 4) \, dt = 4t^3 - 4t + C
step 3
Since the particle is initially at rest, v(0)=0v(0) = 0. Therefore, 0=4(0)34(0)+C    C=00 = 4(0)^3 - 4(0) + C \implies C = 0. Thus, v(t)=4t34tv(t) = 4t^3 - 4t
step 4
Set v(t)=0v(t) = 0: 4t34t=0    4t(t21)=0    t(t1)(t+1)=04t^3 - 4t = 0 \implies 4t(t^2 - 1) = 0 \implies t(t - 1)(t + 1) = 0
step 5
Solve for tt: t=0,t=1,t=1t = 0, t = 1, t = -1. Since t0t \geq 0, the values are t=0t = 0 and t=1t = 1
Answer
The particle is at rest at t=0t = 0 and t=1t = 1.


 (b) Write an expression for the position x(t)\boldsymbol{x}(t) of the particle at any time t0t \geq 0.
step 1
To find the position x(t)x(t), we integrate the velocity function v(t)=4t34tv(t) = 4t^3 - 4t
step 2
Integrate v(t)v(t): (4t34t)dt=t42t2+C\int (4t^3 - 4t) \, dt = t^4 - 2t^2 + C
step 3
Use the initial condition x(1)=3x(1) = 3 to find CC: 3=(1)42(1)2+C    3=12+C    C=43 = (1)^4 - 2(1)^2 + C \implies 3 = 1 - 2 + C \implies C = 4
step 4
Therefore, the position function is x(t)=t42t2+4x(t) = t^4 - 2t^2 + 4
Answer
The position of the particle at any time t0t \geq 0 is x(t)=t42t2+4x(t) = t^4 - 2t^2 + 4.


 (c) Find the total distance traveled by the particle from t=0t=0 to t=2t=2.
step 1
The total distance traveled is the integral of the absolute value of the velocity function v(t)=4t34tv(t) = 4t^3 - 4t from t=0t=0 to t=2t=2
step 2
Find the points where v(t)=0v(t) = 0 within the interval [0,2][0, 2]: 4t(t21)=0    t=0,t=14t(t^2 - 1) = 0 \implies t = 0, t = 1
step 3
Split the integral at these points: 024t34tdt=014t34tdt+124t34tdt\int_0^2 |4t^3 - 4t| \, dt = \int_0^1 |4t^3 - 4t| \, dt + \int_1^2 |4t^3 - 4t| \, dt
step 4
Evaluate the integrals: - For 0t10 \leq t \leq 1, 4t34t04t^3 - 4t \leq 0, so 4t34t=(4t34t)=4t4t3|4t^3 - 4t| = -(4t^3 - 4t) = 4t - 4t^3. - For 1t21 \leq t \leq 2, 4t34t04t^3 - 4t \geq 0, so 4t34t=4t34t|4t^3 - 4t| = 4t^3 - 4t
step 5
Integrate each part: 01(4t4t3)dt=[2t2t4]01=21=1\int_0^1 (4t - 4t^3) \, dt = \left[2t^2 - t^4\right]_0^1 = 2 - 1 = 1, 12(4t34t)dt=[t42t2]12=(168)(12)=8+1=9\int_1^2 (4t^3 - 4t) \, dt = \left[t^4 - 2t^2\right]_1^2 = (16 - 8) - (1 - 2) = 8 + 1 = 9
step 6
Add the results: 1+9=101 + 9 = 10
Answer
The total distance traveled by the particle from t=0t=0 to t=2t=2 is 1010 units.
Key Concept
Integration of acceleration and velocity functions
Explanation
To find when the particle is at rest, we integrate the acceleration to get velocity and solve for when it is zero. To find the position, we integrate the velocity and use initial conditions. To find the total distance, we integrate the absolute value of the velocity over the given interval.

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