# d=rt help?

At 10 : 45 A.M. two people leave their homes that are 6 miles apart and begin walking toward each other. If one person walks at a rate that is 2 mph faster than the other and they meet after 1.5 hours, how fast was each person walking?

### 5 Answers

- KrishnamurthyLv 73 weeks ago
At 10 : 45 A.M. two people leave their homes that are 6 miles apart

and begin walking toward each other.

If one person walks at a rate that is 2 mph faster than the other

and they meet after 1.5 hours,

how fast was each person walking?

Let the two walking speeds be x & (x + 2) mph

Use d = rt

15 = 1.5 x + 1.5(x + 2)

15 = 1.5x +1.5x + 3

12 = 3x

x = 12/3 = 4

One person is walking at 4 mph and the other at 6 mph.

- PinkgreenLv 73 weeks ago
Let Va, Vb mph be the walking speed of person A, B

respectively, Va=Vb+2. Then

6/(Va+Vb)=1.5

=>

6/(Vb+2+Vb)=1.5

=>

3/(Vb+1)=1.5

=>

Vb=1 mph

Va=1+2=3 mph.

- ComoLv 73 weeks ago
:-

Let persons be A and B

Let speed of A = x mph

Speed of B = x + 2 mph

Relative speed of B to A = 2x + 2 mph

Meet after 1•5 hours

Distance travelled by B = 1•5 ( 2x + 2 ) = 6 miles

3 x + 3 = 6

x. = 1

Speed of A = 1 mph

Speed of B = 3 mph

- Ian HLv 73 weeks ago
1mph and 3mph relative speed is 4 mph.

6 miles in 1.5 hours is 4 mph

If you want to use algebra write v + v + 2 = 6/1.5

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- 冷眼旁觀Lv 63 weeks ago
Method 1:

Let r mph be the speed of the slower one.

Then, speed of the faster one = (r + 2) mph

Total distance taken (in miles):

1.5r + 1.5(r + 2) = 6

3r + 3 = 6

3r = 3

r = 1

r + 2 = 3

Their speeds are 1 mph and 3 mph respectively.

====

Method 2:

Let x mph and y mph be their speeds respectively, where y > x

y = x + 2 …… [1]

1.5x + 1.5y = 6 …… [2]

Substitute [1] into [2]:

1.5x + 1.5(x + 2) = 6

3x + 3 = 6

x = 1

Substitute x = 1 into [1]:

y = 1 + 2

y = 3

Their speeds are 1 mph and 3 mph respectively.