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

Answer:

$T=35.625sec$

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

$T=\frac{Distance D}{Relative\ speed V_r}$

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

$T=\frac{ D}{ V_r}$

$T=\frac{ 285}{ 8}$

$T=35.625sec$