Two particles are moving in straight lines. The displacement (in meters) of particle 1 is given by the function e^ (4cos (t)), where t is in seconds. The displacement (in meters) of particle 2 is given by the function - (t^3) / (3) - (t^2) / (2) + 2, where t is in seconds. Find the first positive time at which the particles have (approximately) the same velocity.

A.) t = 1.569 seconds

B.) t = 0 seconds

C.) t = 2.366 seconds

D.) t = 0.588 seconds

E.) t = 1.011 seconds

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.