Position x(t) for Particle Moving at Acceleration a(t)=6t-2

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SUMMARY

The position of a particle moving along the x-axis is defined by the equation x(t) = t^3 - t^2 + 4t + 6, derived from the acceleration function a(t) = 6t - 2. The constants C1 and C2 were determined using the conditions that the velocity is 25 when t = 3 and the position is 10 when t = 1. The integration of acceleration to find velocity and then position is crucial in solving this problem. This analysis demonstrates the application of calculus in determining the motion of a particle.

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The acceleration of a particle moving along
the x-axis at time t is given by a(t) = 6t - 2.
If the velocity is 25 when t = 3 and the po-
sition is 10 when t = 1, then finnd the position
x(t).
 
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Do you know calculus??Apply the definition of both acceleration wrt velocity & velocity wrt coordinate.
Take care of the conditions which will fix integration constants...

It shouldn't be 2 difficult.

Daniel.
 
the_d said:
The acceleration of a particle moving along
the x-axis at time t is given by a(t) = 6t - 2.
If the velocity is 25 when t = 3 and the po-
sition is 10 when t = 1, then finnd the position
x(t).

Let a(t) be the acceleration defined above, v(t) the velocity, and x(t) the x-axis position. By definition:

v(t) = \int a(t) dt = \int (6t - 2) dt = 3t^2 - 2t + C_1
x(t) = \int v(t) dt = \int (3t^2 - 2t + C_1) dt = t^3 - t^2 + C_1*t + C_2

where C1 and C2 are constants. We determine the latter constants by using values given in the problem statement:

v(3) = 25 = 3*(3)^2 -2*(3) + C_1
C_1 = 4

x(1) = 10 = (1)^3 - (1)^2 + (4)*(1) + C_2
C_2 = 6

\color{red} x(t) = t^3 - t^2 + 4t + 6
 

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