A coin dropped in the lift it takes time 0.5 s to reach the floor when lift is staionary it takes time t when lift is moving up

Physics

1 answer:

BARSIC [14]3 years ago

4 0

Answer:

t₁ > t₂

Explanation:

A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂, t₁ = t₂, t₁ >> t₂ , t₂ > t₁

Solution:

Newton's law of motion is given by:

s = ut + (1/2)gt²;

where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.

u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.

hence:

When stationary:

$s=\frac{1}{2}gt_1^2\\\\t_1^2=\frac{2s}{g} \\\\When\ moving\ with\ acceleration(a):\\\\s=\frac{1}{2}(g+a)t_2^2\\\\t_2^2=\frac{2s}{g+a}$

Hence t₂ < t₁, this means that t₁ > t₂.

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If you notice any mistake in my english, please let me know, because i am not native.