The position of an object is given by x = bt^3 - ct^2 + dt. What is the instanta

Question

The position of an object is given by x = bt^3 - ct^2 + dt. What is the instantaneous acceleration of the object when t = 0.7s? Assume b = 4.1m/s^3, c = 2.2m/s^2, and d = 1.7m/s. The position of an object is given by x = bt^3 - ct^2 + dt. What is the instantaneous acceleration of the object when t = 0.7s? Assume b = 4.1m/s^3, c = 2.2m/s^2, and d = 1.7m/s. The position of an object is given by x = bt^3 - ct^2 + dt. What is the instantaneous acceleration of the object when t = 0.7s? Assume b = 4.1m/s^3, c = 2.2m/s^2, and d = 1.7m/s.

Explanation / Answer

instantaneous accelaration a = d^2x/dt^2 = (d/dx)(3bt^2-2ct+d)

a = 6bt-2c

at t= 0.7s

a = (6*4.1*0.7)-(2*2.2) = 12.82 m/s^2

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