Measuring Acceleration Using A Coin

An elevator is descending with a uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 1.5 m above the floor of the elevator at the time of being dropped. The person observes that the coin strikes the floor after 1 s. What is the acceleration of the elevator to the nearest integer?


The answer is 7.

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4 solutions

Discussions for this problem are now closed

Tanmay Jaiswal
Mar 15, 2014

first we must find the acceleration of the coin in the lift using the equation S=ut+1/2*at^2 where-
S = displacement,
u = initial velocity,
t = time taken and
a = acceleration.
Substituting the values in the formula as u = 0, t = 1, and S=1.5, we can find acceleration to be equal to 3 m/s^2. We know that acceleration due to gravity is equal to 9.8 or the closest integral value 10 m/s^2. The motion of the lift relative to the ground is downwards hence the relative acceleration is found as above to be 3. Knowing the acceleration we can calculate acceleration of the lift using the formula-
relative acceleration = Acceleration on body 1 - Acceleration on body 2 (with +ve sign on the acceleration if in the same direction) . Relative acceleration = 10 - 3.
Using this formula we get acceleration to be equal to 7 m/s^2


nice soloution

Uttam Bhadauriya - 7 years, 2 months ago
Magan Singh
Mar 6, 2014

just observe the motion of coin(inside lift) with respect to lift.

covered distance = (1/2) * acc * (time)^2

1.5 * 2=(10-a)(1)^2

a=7m/s2

What if the lift was ascending?

Thinker Mahmud - 7 years, 2 months ago

same but different way. try 1,5 is 1/2.(a+g).t(sqr)2

Hafizh Ahsan Permana - 7 years, 2 months ago

S=ut+1/2at^{2} u=0 1.5×2=a(1)^{2} a=3 Acc due to gravity= acc. Of coin+acc. Of elevator Acc of elevator=9.8-3 A=6.8=7m/s^{2}

Harpinder Singh - 6 years, 11 months ago

s=ut- 1 2 a t 2 \frac { 1 }{ 2 } a{ t }^{ 2 } u=0 ,so that a=2s/t^2 ; Given, s=1.5 and t=1 .so that, a=3ms^-1.then,g-a= 9.8-3=6.8ms^-2 . 7 is the nearest integer of 6.8 . so that answer is 7ms^-2.

Alejandro Romero
Mar 15, 2014

Movement equation for the coin: y'=y-v * t-0.5 * g * t^2

y'=final position; y=initial position; v=initial speed; g=gravity (~10m/s^2); t=time

Movement equation for the elevator's floor: x'=x-u * t-0.5 * a * t^2

x'=final position; x=initial position; u=initial speed; a=acceleration; t=time

(note: i assume both speed and acceleration negative because we are falling)

Initial conditions:

y=1.5m

x=0m (the floor of the elevator at t=0s is our origin)

v=u (which is the speed of the coin and the elevator had at t=0s)

Final conditions:

when the coin reaches the floor at t=1s, x=y

Let's solve:

                        y-v * t-0.5*g * t^2=x-u * t-0.5 * a * t^2

(we simplify v * t and u * t because are the same and are at both sides)

                         1.5-0.5 * 10 * 1^2=0-0.5 * a * 1^2  

                         1.5-5=-0.5 * a

                        -(1.5-5)/0.5=a

                         a=7 m/s^2

(it's a possitive value because we alredy set the sign in the equation)

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