Expert: Steve Holleran - 4/14/2008

Question
Consider a particle moving according to the velocity function,  v(t)=2a-3exp(-2t) + 2/t+2, 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.  

Answer
Hello Antony,

I hope I've interpreted your equation correctly:


                  v(t) = 2a - 3e^(-2t) + 2 / (t+2)

If so, then I can help with parts a and c, but I don't understand where part b is coming from.  If the particle has a trajectory, shouldn't there be some trig functions involved?  In any case, here's what I could come up with.

a)  d =  20 = INT, t=0 to t=3, [2a - 3e^(-2t) + 2/(t+2) dt]

= INT, 0 to 3, [2a dt] -3 INT, 0 to 3,[e^-2t  dt] + 2 INT, 0 to 3, [dt/(t+2)]

= 2a[t], 0 to 3   - 3 * -1/2 * INT[e^-2t * -2 dt], 0 to 3  + 2 [ln abs(t+2)], 0 to 3

= 6a + 3/2 * [e^-2t]0 to 3 + 2[ln 5 - ln 2]

= 6a + 3/2 [1/e^6 - 1] + 2 ln(5/2)

So a = [20 - 3/2[1/e^6 - 1] - 2 ln(5/2)] / 6 = 3.2772

Then to get the terminal velocity, take the limit of the velocity as  t --> inf.

When you do that, the 3e^-2t and the 2/t+2 both --> 0,

so the "2a" value , 6.55,  is the terminal velocity.



c)  a(t) = v'(t) = -3 * -2e^(-2t) + 2 * -1(t+2)^-2

        = 6 / e^2t - 2/(t+2)^2



When v = 6.55, 6.55 = 6.55 - 3e^(-2t) + 2/(t+2)

so             2/(t+2) = 3e^(-2t) and the only way I could solve this is graphically, getting t = 0.7sec.

Then a(0.7) = 6/e^1.4  - 2/2.7^2 = 1.205 m/sec^2


That's the best I could work out.  Like I said, I don't understand the trajectory question.

Steve