Given:
Acceleration of elevator (a) = 2 m/s² (upward)
Initial velocity of elevator (u) = 0 m/s (since the elevator starts from rest)
Final velocity of elevator (v) = 10 m/s (upward)
Height of coin (h) = 1.5 m

To Find:
Time taken by the coin to reach the floor.

Assumptions:
1. The coin is dropped from rest.
2. The height of the elevator is much greater than the height of the coin, so the effect of air resistance can be neglected.

Concepts Used:
1. Motion equations for uniformly accelerated motion:
- v = u + at
- s = ut + (1/2)at²
- v² = u² + 2as
2. Equivalence of acceleration due to gravity and acceleration of an elevator.

Solution:
Step 1: Determine the time taken by the elevator to reach a speed of 10 m/s.
Using the equation v = u + at, we can find the time taken (t₁) by the elevator to reach a speed of 10 m/s.
Given:
u = 0 m/s
v = 10 m/s
a = 2 m/s²
Using the equation v = u + at, we can rearrange it to find t:
t = (v - u) / a
t₁ = (10 - 0) / 2
t₁ = 5 s

Step 2: Determine the displacement of the elevator during this time.
Using the equation s = ut + (1/2)at², we can find the displacement (s₁) of the elevator during this time.
Given:
u = 0 m/s
t = 5 s
a = 2 m/s²
Using the equation s = ut + (1/2)at², we can rearrange it to find s:
s = ut + (1/2)at²
s₁ = 0 + (1/2)(2)(5)²
s₁ = 0 + (1/2)(2)(25)
s₁ = 0 + 25
s₁ = 25 m

Step 3: Determine the time taken by the coin to reach the floor.
Since the coin is dropped from a height of 1.5 m, its initial velocity (u) is 0 m/s.
Using the equation s = ut + (1/2)at², we can find the time taken (t₂) by the coin to reach the floor.
Given:
u = 0 m/s
s = 1.5 m (negative sign indicates downward direction)
a = -9.8 m/s² (acceleration due to gravity)
Using the equation s = ut + (1/2)at², we can rearrange it to find t:
t = (-u ± √(u² - 2as)) / a
Since the coin is dropped, its initial velocity (u) is 0, so the equation becomes:
t₂ = ± √(u² - 2as) / a
t₂ = ± √(0² - 2(-