A particle moves with acceleration function a(t) = 5 + 4t − 2t^2 4.?

Question

Its initial velocity is v(0) = 2m/s and initial position is s(0) = 10 m. Find its position after t seconds.

Answer 1

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a(t) = 5 + 4t -2t^2

a(t) = dv/dt

dv/dt = 5 + 4t -2t^2

dv = (5 + 4t - 2t^2) dt

v = 5t + 2t^2 - (2/3)t^3 + c_1

v(t = 0) = 2 m/s

v = 5*0 + 2*0^2 - (2/3)*0^3 + c_1 = 2 m/s

c_1 = 2 m/s

v = 5t + 2t^2 - (2/3)(t^3) + 2

v = ds/dt

ds/dt = 2 + 5t + 2t^2 - (2/3)(t^3)

ds = [2 + 5t + 2t^2 - (2/3)(t^3)] dt

s = 2t + (5/2)t^2 + (2/3)t^3 - (1/6)(t^4) + c_2

s(t = 0) = 2t + (5/2)t^2 + (2/3)t^3 - (1/6)(t^4) + c_2 = 10

s(t = 0) = 2*0 + (5/2)*0^2 + (2/3)*0^3 - (1/6)(0^4) + c_2 = 10 m

c_2 = 10 m

s = 2*t + (5/2)*t^2 + (2/3)*t^3 - (1/6)(t^4) + 10

s = 10 + 2*t + (5/2)*t^2 + (2/3)*t^3 - (1/6)(t^4)