Question

On a rectangle with length 100 m, width 50m and 2 vehicles stationed together. Find the time that they meet given that car A travels 5 m/s and leaves 3 seconds before car B, and car B is traveling at 3 m/s in the opposite direction. Can you create a generic equation from the previous scenario

Answer 1

`T=35.625sec`

Step-by-step explanation:

From the question we are told that:

Length
`L=100 m`

Width
`W=50m`

Velocity of Car A
`V_A=5m/s`

Velocity of Car B
`V_B=3m/s`

Distance traveled by car A before car B moves

`d_l=5*3`

`d_l=15`

Therefore total distance traveled at same time interval

`D=total\ distance-d_l`

Where

Total distance=Perimeter of rectangle

`P=2(L+B)`

`P=2(100+50)`

`P=300`

Therefore

`D=total\ distance-d_l`

`D=300-15\\D=285m`

Generally the equation for time taken to meet is mathematically given by

`T=(Distance D)/(Relative\ speed V_r)`

Where

Relative speed = Speed of car A +Speed of car B

`V_r=V_A+V_B`

`V_r=5+3`

`V_r=8m/s`

Therefore the time taken to meet

`T=( D)/( V_r)`

`T=( 285)/( 8)`

`T=35.625sec`