The answer is 30s.

there is a formula for this type of questions - [(αβ)/2(α+β)] * t2 = s

where α = Maximum acceleration t = time

β = Maximum retardation s = distance

Solution: Let (D) represent the distance covered during acceleration.

Then (1500 - D) would be the distance covered during negative acceleration.

Let t (1) represent the time lapse during acceleration, and t (2) represent the time lapse during negative acceleration. The total time T would be [t (1) + t (2)]; and T would be the measurement of time that we are seeking.

The definition of acceleration is delta V / delta t. Since we are taking that it started from rest, then the velocity at t = t (1) would be: V (1) = 5 t (1).

Since we are taking it that the cycle came to rest at the end, V (1) = 10 t (2).

Substitution axiom tells us that 5 t (1) = 10 t (2). This reduces to t (2) = half t (1) which makes total sense since retardation is double the acceleration.

Using kinematic equation [d = V (i) t + 1 / 2 a t^2] for the first leg, we get:

D = 1/2 (5) [t (1)]^2 and for the second leg: (1500 - D) = [{5 t (1)} {t (2)}] + 1/2 (-10) [t (2)^2]

Substitution for t (2) gives us: (1500 - D) = [{5 t (1)} {t (1)/2}] + 1/2 (-10) [{ t (1)}/2]^2

Substitution for D gives us: [1500 - (1/2) (5) {t (1)}^2] = [{5 t (1)} {t (1)/2}] + 1/2 (-10) [{ t (1)}/2]^2

Remove brackets and solve for t (1). We get:

1500 - (5/2) [t (1)^2)] = 5/2 [t (1)^2] - (5/4) [t (1)^2]

6000 / 15 = t (1)^2 → t (1) = 20 seconds

Then t (2) = 20/2 = 10 seconds. Total trip = 30 seconds.