Answer:
t = 0second (B)
Step-by-step explanation:
Given the displacement of the first particle to be r = e^(4cos(t))
Since velocity = Change in displacement/time
Differentiating the displacement with respect to time (t), we have
Velocity = dr/dt = -4sinte^(4cos(t))
Velocity of the first particle is -4sinte^(4cos(t))... (1)
Similarly for particle 2,
If its displacement r= -(t^3)/(3) - (t^2)/(2) + 2
dr/dt = -3t²/3-2t/2
dr/dt = -t²-t
Velocity of the second particle is
-t²-t...(2)
First positive time at which the particles will have(approximately) the same velocity will be when the velocity of the first particle equals that of the second particle i.e
-4sinte^(4cos(t)) = -t²-t
Equating this velocity equations to 0, we will have
4sinte^(4cos(t)) = -t²-t = 0
Therefore,
4sinte^(4cos(t)) = 0... (3)
-t²-t = 0 ... (4)
From equation 1,
4sint = 0
Sint = 0
t = 0second
From equation 2,
-t²-t = 0
-t(t+1) = 0
-t = 0 and t = -1second
t = 0second
Therefore the first positive time will be 0second. -1second will be ignore since it is an negative time.
