How Fast is a Clock Moving if It Runs Four Times Slower?

[ZedCar]

A moving clock is observed to run four times more slowly than an identical
stationary clock. What is its speed relative to the frame of the stationary clock?

I'm wondering which equation should I be using for this?

Thanks

 

[CompuChip]

What equations do you know of that might be relevant here?
Which of the quantities in those equations are given (and what is the one you want to calculate)?

 

[ZedCar]

I've been using

t = t0 / sqrt(1 - v^2/c^2)

t = 4t / (1 - v^2/c^2)^0.5

(1 - v^2/c^2)^0.5 = 4t/t

1 - v^2/c^2 =16

-15c^2 = v^2

v = (-15c^2)^0.5

 

[The1337gamer]

This is what i think it is:

delta t' = delta t/sqrt(1-(v/c)^2)

now
delta t = proper time, time interval in moving frame as measured by the moving frame
delta t' = time interval of moving frame as measured by rest frame.

So delta t' = 4 delta t, as the time interval of the moving clock is measured 4 times slower than stationary clock.

Therefore 4 = 1/sqrt(1-(v/c)^2)

Then you should be able to solve and find v as a fraction of c.

 

[ZedCar]

Thank you.

I'm now getting answer of ((15/16)^0.5)c