At time 𝑡≥ 0 , a particle moving in the xy-plane has velocity vector given by 𝑣(𝑡)=〈3,2−𝑡2〉 . If the particle is at the point (1,12) at time 𝑡= 0 , how far is the particle from the origin at time 𝑡= 1 ?Responses2.3042.3043.1073.1074.2094.2095.310
Question
At time 𝑡≥ 0 , a particle moving in the xy-plane has velocity vector given by 𝑣(𝑡)=〈3,2−𝑡2〉 . If the particle is at the point (1,12) at time 𝑡= 0 , how far is the particle from the origin at time 𝑡= 1 ?Responses2.3042.3043.1073.1074.2094.2095.310
Solution
To solve this problem, we first need to find the position function of the particle. The position function is the integral of the velocity function.
The velocity function is given by 𝑣(𝑡)=〈3,2−𝑡²〉.
We can integrate this function with respect to t to find the position function:
∫𝑣(𝑡) dt = ∫〈3,2−𝑡²〉 dt = 〈3t, 2t - t³/3〉 + C
We know that at time t=0, the particle is at the point (1,12). We can use this information to solve for the constant of integration C:
(1,12) = 〈30, 20 - 0³/3〉 + C => C = (1,12)
So the position function is:
r(t) = 〈3t, 2t - t³/3〉 + (1,12) = 〈3t+1, 2t - t³/3 + 12〉
At time t=1, the position of the particle is:
r(1) = 〈31+1, 21 - 1³/3 + 12〉 = 〈4, 13 - 1/3〉 = 〈4, 12 2/3〉
The distance of the particle from the origin at time t=1 is the magnitude of the position vector r(1):
|r(1)| = sqrt((4)² + (12 2/3)²) = sqrt(16 + 160.44) = sqrt(176.44) = 13.28
So, the particle is approximately 13.28 units from the origin at time t=1.
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