Final answer:
a. To show that the velocity is never zero, we can prove that the expression 4e^(2t) + 6e^(-3t) is always greater than zero. b.
To find the acceleration when t = ln 2, we differentiate the equation for velocity and substitute t = ln 2. c. To find the displacement of the particle from O when t = 2, we integrate the equation for velocity and substitute t = 2.
Step-by-step explanation:
a. Show that the velocity is never zero.
To show that the velocity is never zero, we can prove that the expression 4e^(2t) + 6e^(-3t) is always greater than zero. Let's assume v = 4e^(2t) + 6e^(-3t) is equal to zero. Rearranging the terms, we get 6e^(-3t) = -4e^(2t).
Taking the natural logarithm of both sides, ln(6) - 3t = ln(-4) + 2t. However, the natural logarithm of a negative number is undefined, so this equation has no solution. Therefore, the velocity is never zero.
b. Find the acceleration when t = ln 2.
To find the acceleration when t = ln 2, we need to differentiate the equation for velocity with respect to time. The given equation for velocity is v = 4e^(2t) + 6e^(-3t). Taking the derivative, we get a = 8e^(2t) - 18e^(-3t). Substituting t = ln 2 into the equation, we get a = 8e^(2ln 2) - 18e^(-3ln 2).
Simplifying further, we have a = 8(2^2) - 18(2^(-3)). Evaluating the exponents, we get a = 32 - 9/4. Therefore, the acceleration when t = ln 2 is 32 - 9/4 ms^-2.
c. Find, to the nearest metre, the displacement of the particle from O when t = 2.
To find the displacement of the particle from O when t = 2, we need to integrate the equation for velocity with respect to time. The given equation for velocity is v = 4e^(2t) + 6e^(-3t). Integrating the equation, we get the displacement function d = 2e^(2t) - 2e^(-3t) + C, where C is the constant of integration. Since the particle passes through O at t = 0, we can substitute d = 0 into the equation and solve for C. 0 = 2e^(2*0) - 2e^(-3*0) + C. Simplifying, we have 0 = 2 - 2 + C. Therefore, C = 0.
Substituting t = 2 into the displacement function, we get d = 2e^(2*2) - 2e^(-3*2) + 0. Evaluating the exponents, we have d = 2e^4 - 2e^(-6). Calculating the values, we get d ≈ 175 metres. Therefore, the displacement of the particle from O when t = 2 is approximately 175 metres.