Question

a(t) = t^2 - 4t + 6, s(0) = 0, s(1) = 20      A particle is moving with the given data. Find the position of the particle.

Accepted Answer

a(t)=t^2-4t+6
a(t)=v'(t)
v(t)=inta(t)dt
v(t)=int(t^2-4t+6)dt 
v(t)=t^3/3-4(t^2/2)+6t+c_1
v(t)=t^3/3-2t^2+6t+c_1
s(t)=intv(t)dt
s(t)=int(t^3/3)-2t^2+6t+c_1)dt 
s(t)=(1/3)(t^4/4)-2t^3/3+6t^2/2+c_1t+c_2
s(t)=t^4/12-(2t^3)/3+3t^2+c_1t+c_2
Let's find constants c_1 and c_2 . given s(0)=0 and s(1)=20
s(0)=0=c_2
s(1)=20=1/12-2/3+3+c_1
20=(1-8+36)/12+c_1
20=29/12+c_1
c_1=20-29/12=211/12
s(t)=t^4/12-2t^3/3+3t^2+211/12t

Answer

Integrating for velocity: v(t) = (1/3)t^3 - 2t² + 6t + C   Integrating for position: s(t) = (1/12)t^4 - (2/3)t^3 + 3t² + Ct + D   Using the 2nd given value 20 = (1/12) - (2/3) + 3 + C + D   Solving D D = -C - 17.58   Back to this equation: s(t) = (1/12)t^4 - (2/3)t^3 + 3t² + Ct + D  s(t) = (1/12)t^4 - (2/3)t^3 + 3t² + Ct - C - 17.58   Using the first given value: 0 = 0 - 0 + 0 + 0 - C - 17.58  C = -17.58   Back this equation again: s(t) = (1/12)t^4 - (2/3)t^3 + 3t² + Ct + D  s(t) = s(t) = (1/12)t^4 - (2/3)t^3 + 3t² - 17.58t - C - 17.58  s(t) = (1/12)t^4 - (2/3)t^3 + 3t² - 17.58t + 17.58 - 17.58  s(t) = (1/12)t^4 - (2/3)t^3 + 3t² - 17.58t

Math

`a(t) = t^2 - 4t + 6, s(0) = 0, s(1) = 20` A particle is moving with the given data. Find the position of the particle.

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