Two asteroids in the Oort Cloud are on a direct collision path with each other. One asteroid is traveling at 750 km/hr while the other travels at 1050 km/hr.

Question

If the two asteroids are 3.6✕10^6 meters away from each other, how many meters will the faster asteroid have traveled at the time of impact?

A) 3.6✕10^6

B) 2.1✕10^6

C) 1.8✕10^6

D) 1.5✕10^6

Answer 1

If they were moving at the same rate, they'd meet at the middle of 3.6x10^6 m: 1.8x10^6. But since one is faster, that one will cover more distance, so it has to be above 1.8x10^6 (and it can't be 3.6x10^3, since both objects are moving): therefore, 2.1x10^6 is the answer, assuming one of the choices offered is correct.

Answer 2

We need to set up an equation to find the time of collision.

convert the 3.6*10^6m to km

3.6*10^6 m / 1000m/km =3.6*10^3 km

since they are both approaching each other, they are both positive and added to one another.

750t + 1050t = 3600 km

t is time in terms of hours

1800t=3600

t = 2 hours

time*velocity= distance

2 hr * 1050km/hr = 2100 km

in meters 2100 km *1000m/km = 2.1 *10^6 m B)