The acceleration of a particle is directly proportional to the square of the tim

Question

The acceleration of a particle is directly proportional to the square of the time t. When t = 0, the particle is at x = 36 ft. Knowing that at t = 9s, x = 144 ft and v = 27 ft/s, express x and v in terms of t.

Explanation / Answer

The acceleration of a particle is directly proportional to the square of the time t. the problem is actually much easier. I misread and assumed it was proportional to the SQUARE ROOT of t, rather than the square. I therefore have corrected below. The acceleration is directly proportional to the square of t, so in other words if k is an arbitrary constant: a = k*t^2 Integrating with respect to t to get v: v = Int(a, t) = 1/3 * k * t^3 + C Where C is a constant of integration. Integrating again to get an expression for the distance x: x = Int(v, t) = 1/12 * k * t^4 + C*t + D Where D is a second constant of integration. Now you have two equations with three unknowns (eqns for v and x) Use your initial conditions to solve for k, C, and D. 1) t=0, x=24 m 24 = D -> D = 24 m 2) t=6, x=96 m 96 = 1/12*k*(6^4) + C*6 + 24 --> C=-6 3) t=6, v=18 18 = 1/3*k*(6^(3)) - 6 --> k=1/3 Then you can write v and x in terms of t. Note that the carrot is an exponent. I did the math in my head for the last part so you might need to check. When t = 0, the particle is at x = 36 ft. Knowing that at t = 9s, x = 144 ft and v = 27 ft/s, express x and v in terms of t.

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