Explanation for Question 27 From the Math (Calc) Section on the 2018 May Sat
Question 27 says a signal from a spacecraft orbiting mercury travels 2 to earth at a speed of three times 10 to the eighth meters per 3 second. If the distance between earth and the spacecraft is 2.02 4 to one times 10 to the eighth kilometers, 5 which of the following is closest to the number of minutes. 6 It will take for a signal from the spacecraft to reach earth. 7 So there's a couple of steps. 8 This first, we know that speed is equal to distance 9 over time, but here we're given the speed 10 and we're given the distance. So we need to solve for time. 11 So we can just change this equation around by multiplying both sides by time 12 and dividing both sides by seat, by speed, 13 to find that time is equal to distance over 14 speed. So if we can plug in the distance and the speed, 15 we can figure out what the Titus, which is what we're looking for. 16 However, we first have to actually convert this because the distance is given in 17 kilometers and we need it in meters. 18 So the distance we're given is equal to 19 2.02 to one times 10 to the eighth 20 kilometers. And to get that to meters, 21 we have to multiply that by 1000 and our new distance 22 is 2.02 to one times 10 to the 11 kilometers. 23 So now we can plug in our distance and our speed and we'll get 24 a time. So the distance is 2.02 to one times 10 25 to the 11th, our speed, 26 or I'm getting from right here. That's three times 10 to the eighth and 27 that's equal to 674.03, 28 but this is giving us the number of seconds that it will take. 29 And we need to convert this to minutes. So to do that, 30 I'm going to divide this number by 60. And that gives me 11.23, 31 which is closest to 11 and therefore.