Question

5. (10 points) To find the motion of the particle we need to solve the following integrals (t-1) (t-2) 2 (t-1) dt

Answer 1

x(t) = integral 3/(t - 1)(t - 2) dt - (1) 1/(t - 1)(t - 2) = A/(t - 1) + B/(t - 2) doublerightarrow 1/(t - 1)(t - 2) = A(t - 2) + B(t - 1)/(t - 1)(t - 2) doublerightarrow A(t - 2) + B(t - 1) = 1 --- (2) put t = 1 in the eq. No (2) then we will get doublerightarrow A(1 - 2) + B(1 - 1) = 1 doublerightarrow -A + 0 = 1 therefore A = -1 Put P = 2 in the eq. No 92), then we will get doublerightarrow A(2 - 2) + B(2 - 1) = 1 doublerightarrow 0 + B(1) = 1 therefore B = 1 x(t) = integral 3/(t - 1)(t - 2) dt = 3 integral[-1/t - 1 + 1/t -2] dt 3[- integral 1/t -1 dt + integral 1/t - 2 dt] = 3[- log(t - 1) + log(t - 2)] + C [therefore integral 1/x dx = log x + C] = 3[log(t - 2) - log(t - 1)] + C therefore x(t) = integral 3/(t - 1)(t - 2) dt = 3[log(t - 2) - log(t - 1)] + C