Question

A particle moves along the x-axis with acceleration at any time t given as a(t)=3t²+4t+6. If the particle's initial velocity is 10 and its initial position is 2 , what is the position function?

Answer 1

The position function of the particle, given its acceleration function a(t) = 3t² + 4t + 6, initial velocity of 10 m/s, and initial position of 2 m, is x(t) = ¼ t´ + ½ t³ + 3t² + 10t + 2.

Step-by-step explanation:

To determine the position function of a particle given its acceleration function a(t) = 3t² + 4t + 6, initial velocity v0 = 10 m/s, and initial position x0 = 2 m, we integrate the acceleration function with respect to time to find the velocity function, v(t), and then integrate v(t) to obtain the position function, x(t).

The velocity function is found by integrating the acceleration function:

v(t) = ∫(3t² + 4t + 6)dt = t³ + 2t² + 6t + C

To find the constant of integration C, we use the given initial velocity:

v(0) = 10 = 0³ + 2×0² + 6×0 + C ⇒ C = 10

So the velocity function is v(t) = t³ + 2t² + 6t + 10.

Now, to find the position function, we integrate the velocity function:

x(t) = ∫(t³ + 2t² + 6t + 10)dt = ¼ t´ + ½ t³ + 3t² + 10t + K

Using the initial position to find the constant K:

x(0) = 2 = ¼×0´ + ½×0³ + 3×0² + 10×0 + K ⇒ K = 2

Thus, the position function is x(t) = ¼ t´ + ½ t³ + 3t² + 10t + 2.