A dog is 100 feet away from the tree, and cat is 75 feet away from the tree?

100/24= 4.166667 seconds

75/4.166667 = 18 feet per second

so atleast 18 feet per second. But to sound like a genius u could put that it would have to be faster cos the cat would have to react first, nd the dog would already be closer.

Solution:

Obviously, the cat has to reach the tree to escape from the dog. I will assume that the positions of the dog and the cat are such that they lie along a straight line that includes the tree.

Solution:

for the dog:

distance = rate x time

100 ft = 24 ft/sec x time

time = 100 ft/(24 ft/sec)

time = 4 1/6 sec or 4.17 sec

In other words, the cat has to reach the tree in 4 1/6 sec in order to climb up and escape the dog. Hence,

for the cat:

distance = rate x time

75 ft = rate x 4 1/6 sec

rate = 75 ft/(4 1/6 sec)

rate = 18 ft/sec

Answer: The cat has to run at the rate of

of 18 ft/sec.

Hope I help.

teddyboy

The dog is 100ft from the tree. If it can run at 24 ft per seconds it will reach the tree in 100/24 seconds.

The cat needs to beat the dog to the tree so it has to travel 75 feet in 100/4 seconds. Speed = distance / time

So

speed = 75 / (100/24) = 75 * 24 / 100 = 18

The cat has to run at least just over 18 ft per second (if it travels at exactly this speed, both dog and cat arrive in a dead heat and the dog therefore catches the cat).

If the cat can run at 24 feet/sec, the dog should not be able to catch her. Why did I assume the cat was female?

*EDIT*

Oh I didn't realise the cat was planning to climb the tree. In that case those who are saying >18ft/s will be correct. Whereever the cat is wrt the dog initially, it can always escape by running at >18 ft/s toward the tree. That is if the dog is aiming toward where the cat is running. If the dog aims directly toward the cat at all times, or if the dog tries to intercept the cat anywhere before the tree, then the cat could afford to take some speed off. She could run on 3 legs and still escape.

Let's say the cat is at some location where CTD = θ initially, and the cat aims directly for the tree at 18 m/s. If the dog aims directly for T, it would intercept the cat at T. Now let's suppose the dog were able to intersept the cat at some point along CT, say P. That means both the dog and cat would be at P at the same time. And if they both continued along PT, the dog would be there first. So if the dog can travel in a non straight path DPT and get to T before the cat, then he would get there even faster if he travelled directly along DT. But since travelling along DT allows him to reach T in the same time as the cat, there is no point P along CT where the dog can intercept the cat if the cat travels at 18 ft/s.

If they both run at the same speed, the cat will escape, since it's 25 feet closer to the tree than the dog... This is assuming the cat hasn't been de-clawed, and can't climb the tree.

The dog is on a 3 feet leash. The cat can run, walk or just stay there.

Work of the dog=mad

um...okay

in 4.17 "rounded up...-.16666 infinity" seconds the dog reaches the tree

with the same speed the cat can make the tree in 3.125

ahh crap in 4.16666666666666667 seconds the cat reaches the tree at a speed of

18 feet per second

so anyspeed over 18 fps... unless the cat is a ninja...then the cat used a smoke bomb and makes it to the tree in under 1 second.

Exactly 1 ft/sec faster than the dog.

about 18 ft/sec