going for a bike ride problem

Question

Mike's house is 100 km to the east of Tom's house. At 8:00am, Mike and Tom start biking from their respective houses. Mike bikes west at 1 km/h and Tom bikes south at 5 km/h. If they continue their respective courses, how long after they start will they be closest to each other and that is the distance between them at that time?

t=

d=

Answer 1

(a)

Mike's house is 100 km to the east of the Tom's house

We know that Speed = Distance / time.

Distance = Speed * time

 

Let t be the time

Mike travels at a speed of 1 km/hr in the west direction

Distance = 1*t = t km

Now the distance between tom's house and Mike is (100 - t) km

Tom travels at a speed of 5 km/hr in the south direction

Distance = 5 * t = 5t km

 

Apply Pythagoras rule,

d² = (100 - t)² + (5t)²

d² = 10000 + t² - 200t + 25t²

d² = 26t² - 200t + 10000

 

To find the minimum distance, Differentiate above expression with respect to t and equate d' = 0

2d d' = 52t - 200

52t - 200 = 0

t = 200 / 52

t = 3.846 hours

t = 3 hours : 0.846 * 60 min

t = 3 hours 50.76 mins

Therefore time taken by Tom and Mike to be closest to each other is 3 hours and 50.76 mins.

 

Answer 2

(b)

Distance between tom and mike is d² = 26t² - 200t + 10000   (From (a))

Time taken after 8:00am to travel a distance such that they are closest to each other is 3hours 50.76 mins.

t = 3.846

Distance between tom and mike during time t = 3.846 is d² = 26(3.846)² - 200(3.846) + 10000

d² = 9615.38

d = 98.05 km

Therefore the minimum Distance between Tom and Mike is 98.05 km.