a hare runs 50 ft/sec and a tortoise who is 1000 ft ahead of the hare runs 0.25 ft/sec. Assuming they both start at time t=0, what will be the value of t when the hare finally catches up?
When will the hare catch the tortoise?
The Tortoise approach the Hare with rate (50-.25) ft/sec.
The distanse between them is 1,000 ft.
According to D=R×T formula T=D/R=1,000/(50-.25) sec
Velocity of the hare, = V(h)
Velocity of tortoise, = V(T)
The time when they are even with each other they will have covered the same distance.
Distance, D of the hare, = V(h)t Distance, D of the tortoise, = V(T)t + 1000 ft or
V(h)t = 1000 ft+ V(T)t, solve for t,= 1000ft /(v(h)-v(T)), = 1000 ft/ (50ft/sec - 0.25 ft/sec) = 20.1 sec
We are given three pieces of information. The hare's speed, the tortoise's speed, and the head start that the tortoise has. The units are feet/second for speed and feet for distance. Time will then be in seconds.
The principle is DISTANCE = RATE • TIME, or D = R • T
The rate is the same as the speed.
For the hare, DHare = 50 • T
That is the distance (feet) the hare travels in t seconds.
For the tortoise, DTortoise = .25 • T + 1000
The tortoise gets 1000 added to the distance equation because of the 1000 foot head start.
They well be at the same distance from the starting line when those equations are equal to each other.
50 • T = .25 • T + 1000
Subtract .25 • T from each side:
50 • T - .25 • T = .25 • T + 1000 - .25 • T
49.75 • T = 1000
Divide both sides by 49.75:
T = 1000 / 49.75
T = 20.1 seconds
Hare = 50t
Tortoise = 0.25t + 1000
You want when Hare = Tortoise, so:
50t = 0.25t + 1000
49.75t = 1000
t = 20.10 seconds