A particle moves along the x-axis so that its position as a function of time is

Question

A particle moves along the x-axis so that its position as a function of time is given by the equation x = 4t^6 - 4t^3 +11t + 14 (a) Write the particle?s velocity as a function of time. (Symbolic: Use t for time.) V = [ ] (b) Write the particle?s acceleration as a function of time. (Symbolic: Use t for time.) A = [ ] The velocity of a ball moving along a straight road is given by v(t) = A.t^4 - B.t^2 + C where A, B, and C are constants. At t = 0 the ball is located at the origin, x = 0. Only symbols A, B, C, and t plus numbers can appear in your answer. To see your answer in standard math notation, click on the eye after the answer box. For help with symbolic programming click link at the bottom of the question. (a) Find an expression for the acceleration, a(t) = [] (b) Find an expression for the position, x(t) = [ ] (c) Suppose that the ball is located at x = D (with D non-zero) at time t = 0. Would this change the expression for acceleration? .No .Yes (d) Suppose that the ball is located at x = D (with D non-zero) at time t = 0. Would this change the expression for position? .Yes .No

Explanation / Answer

1.       x(t) = 4t6-4t3 +11t+14

velocity = dx/dt

v(t) = 24t5 - 12t2 +14

acceleration = d2x/dt2 = 120t4 - 24t

2. v (t) = At4 - Bt2 + C

acceleration = dv/dt

acceleration = 4At3 -2Bt

and position is integration of velocity

x = At5/5 - Bt3/3 +Ct + D

where D = constant of integration

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