You and a friend live on opposite ends of a long, straight street and agree to meet somewhere on the street between your homes. You leave promptly at 1pm, traveling at 1.8 m/s toward your friend’s house. Your friend, late as always, doesn’t leave until 1:10, and travels toward your house at 2.1 m/s. If you meet at a point 630 m from your friend’s house, what’s the distance between your house and your friend’s house?

s = 2250 m = 2.25 km

Explanation:

First we need to find the time traveled by the friend. So, we use:

s₁ = v₁ t₁

where,

s₁ = distance traveled by friend = 630 m

v₁ =  speed of friend = 2.1 m/s

t₁ = time taken by friend = ?

Therefore,

630 m = (2.1 m/s)(t₁)

t₁ = 300 s

Since, you traveled 10 mintes = 600 s more than the friend. Hence,

s₂ = v₂ t₂

where,

s₂ = distance traveled by friend = ?

v₂ =  speed of you = 1.8 m/s

t₂ = time taken by you = 300 s + 600 s = 900 s

Therefore,

s₂ = (1.8 m/s)(900 s)

s₂ = 1620 m

Now, the total distance between house will be:

s = s₁ + s₂ = 630 m + 1620 m

s = 2250 m = 2.25 km

your answer is c! hope this : )

conservation of mass

mass before something happens = mass afterwards