[Video] Q14: L=s√1-v^2/c^2when the speed of an object approaches the speed of light, its length as seen by an observer changes. when the object is stationary relative to an observer, its length is s, and when the same object is moving at a speed of v relative to the observer, its length is l. the formula above expresses l in terms of s, v, and c, the speed of light. which of the following gives the speed of the object in terms of the other quantities?

Explanation for Question 14 From the Math (No Calc) Section on the 2020 March Sat

So for number 14 here, the answer key says that the correct answer is 2 a, so let's see how we can land on a and what mistakes will 3 possibly drive us to choose the incorrect answers of B, 4 C or D. And so for 14, 5 right, we have this equation here and they've told us the names of all 6 the variables, but in reality, these names of what the variables are and what, 7 um, what kind of physics principles that's grounded in, 8 right? Being the speed of light. That's not really important. 9 All we need to know is which of the following gives the speed of 10 an object, the speed of the object in terms of the other quantities, 11 all that really means is that they want you to take this equation and 12 they want you to solve it for V. So basically we're not changing anything 13 around, we're just maneuvering stuff or putting things in different positions, 14 but nothing about this equation is inherently changing, 15 right? 16 So let's rewrite it here where we have L is equal to S square 17 root of one minus V squared, over C squared. 18 And I'm trying to get here. I'm trying to get to V. 19 And so the best thing I can do is start with what is easy 20 and on the outside, right? And that's going to be the S I can 21 start by dividing both sides by us, 22 because that will cancel out this S removing it from this, 23 um, radical that I have. So these S's will cancel, 24 and I'll be left with L over S is equal to the square root 25 of one minus V squared, oversea squared. 26 So at this ...