The velocity-time graph can be obtained by integrating the acceleration-time graph.

From t=0 to t=2s, the acceleration is constant at 15m/s^2. Therefore, the velocity at t=2s can be found by using the equation:

v = u + at

where u is the initial velocity (which is not given), a is the acceleration, and t is the time.

Since the particle starts from rest (u=0), we have:

v = 0 + 15(2) = 30m/s

Therefore, the velocity at t=2s is 30m/s.

Now, we need to find the time at which the velocity becomes equal to the initial velocity. Let's call this time t1.

From t=2s to t1, the acceleration is negative (i.e. the particle is decelerating). Therefore, the velocity-time graph will be a straight line with negative slope.

From t1 to t=4s, the acceleration is zero (i.e. the particle is moving with constant velocity). Therefore, the velocity-time graph will be a straight line with zero slope.

From t=4s to t=5.5s, the acceleration is positive (i.e. the particle is accelerating). Therefore, the velocity-time graph will be a straight line with positive slope.

Since the velocity-time graph is a straight line with negative slope from t1 to 2s, the final velocity at t1 can be found by using the equation:

v1 = v2 + a(t2 - t1)

where v2 is the velocity at t=2s (which we have already found to be 30m/s), a is the acceleration during the deceleration phase, t2 is 2s, and v1 is the final velocity at t1.

Since the acceleration is constant during this phase, we can use the average acceleration:

a = (v2 - v1)/(t2 - t1)

Substituting the values, we get:

-15 = (30 - v1)/(t1 - 2)

Solving for v1, we get:

v1 = 45 - 15t1

Now, we need to find the time at which v1 = u (i.e. the velocity becomes equal to the initial velocity).

Substituting v1 = u, we get:

u = 45 - 15t1

Solving for t1, we get:

t1 = 3s

Therefore, the time at which the velocity becomes equal to the initial velocity is 3s.

Answer: Option (c) 4.5s

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