Question

A boat leaves a dock at 5:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 6:00 PM. How many minutes after 5:00 PM were the two boats closest together

Answer 1

Answer:

After 21.6 minutes after 5:00 pm boats are closest to each other.

Step-by-step explanation:

Speed of the first boat = 20 $\frac{km}{hr}$

Let after time t the distance between the boats is minimum.

Distance traveled by the first boat in time t is = 20 t ------- (1)

Speed of the second boat = 15 $\frac{km}{hr}$

Distance traveled by the second boat in time t is = 15 t ------- (2)

From the diagram

Distance AB = 15 - 15 t --------- (3)

Distance between the boats is at time t is AC. And it is calculated by

$AC^{2} = AB^{2} + BC^{2}$

$AC^{2} = (15 - 15 t) + ( 20 t)^{2}$

Differentiate above equation with respect to t we get,

$AC'^{2} =$ 2 (15 - 15 t) (-15) + 2 (20 t) (20)

$AC'^{2} =$ - 450 + 1250 t

for the distance to be minimum $AC'^{2} =$ 0

⇒ - 450 + 1250 t = 0

⇒ t = $\frac{450}{1250}$

⇒ t = 0.36 hours

⇒ t = 21.6 minutes

Therefore the distance between the boats after 21.6 minutes after 5:00 pm is minimum.