Question – 3: Consider a particle moving according to the velocity function,
v(t) = 2a-3exp(-2t)+2/t+2,for t>0.
(a) If the net distance,d,covered by the particle in the time interval,[0,3],is 20,find the value of a. What is the terminal velocity of the particle?.
(b) Find the expression for the trajectory,y(t), such that, y(2) = 5 . Find the initial position of the particle.
(c) Find the acceleration, a(t), of the particle, for t > 0.
Find the acceleration when the velocity is 6.55 m/s.
The net distance (displacement) is INTEGRAL v(t) dt. Do that integration, and solve for a.
Accelerlation is d/dt of v(t)
y(t)= INT v (dt) . Intial position willbe the constant of integration.
To solve these problems, we'll need to apply different techniques of calculus. Let's work through each part of the question step by step.
(a) To find the value of 'a' and the terminal velocity of the particle, we need to integrate the velocity function, v(t). The net distance covered is the integral of the absolute value of v(t) over the time interval [0,3]. We set this integral equal to 20 and solve for 'a'.
The net distance covered, d = ∫(0 to 3) |v(t)| dt = ∫(0 to 3) |2a - 3exp(-2t) + 2/t + 2| dt.
Using the given integral, you can integrate each term of v(t) individually, keeping in mind the limits of integration [0, 3]. This will yield an expression involving 'a'. Then, set this expression equal to 20 and solve for 'a'.
To find the terminal velocity, we need to evaluate the limit of v(t) as t approaches infinity. The terminal velocity occurs when v(t) approaches a constant value at large values of 't'.
(b) To find the expression for the trajectory, y(t), you need to integrate the velocity function, v(t), with respect to 't'. The trajectory represents the position of the particle as a function of time.
Integrate v(t) to obtain an expression for y(t), and then use the given condition y(2) = 5 to solve for the initial position of the particle.
(c) To find the acceleration, a(t), we need to differentiate the velocity function, v(t), with respect to 't'. This will give us the acceleration of the particle as a function of time.
Differentiate v(t) to obtain an expression for a(t).
To find the acceleration when the velocity is 6.55 m/s, set v(t) equal to 6.55 and solve for 't'. Once you find the time value, substitute it into the expression for a(t) to find the corresponding acceleration.
Remember that differentiation and integration are fundamental operations in calculus. Applying these operations correctly will allow you to solve the given problems.