Two people start biking from the same point. one bikes east at 9 mph, the other south at 28 mph. what is the rate at which the distance between the two people is changing after 10 minutes and after 55 minutes

(a) 31.75 mph

(b) 31.77 mph

Explanation:

(a) For t = 10 min = 10 / 60 = 1 / 6 hour

According to the question,

dx /dt = 15 mph

dy / dt = 28 mph

After 10 minutes

x = 1/6 x 15 = 2.5 miles

y = 1/6 x 28 = 4.67 miles

According to diagram

Differentiate both sides with respect to t.

2D dD/dt = 2 x dx/dt + 2y dy/dt

D dD/dt = x dx/dt + y dy/dt

5.3 dD/dt = 2.5 (15) + 4.67 (28)

dD/dt = 31.75 miles/hour

(b) For t = 55 minutes = 55 / 60 hours

After 55 / 60 hours

x = 55 / 60 (15) = 13.75 miles

y = 55 / 60 (28) = 25.67 miles

According to diagram

Differentiate both sides with respect to t.

2D dD/dt = 2 x dx/dt + 2y dy/dt

D dD/dt = x dx/dt + y dy/dt

29.12 dD/dt = 13.75 (15) + 25.67 (28)

dD/dt = 31.77 miles/hour

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s