physics question, please know what your doing?

Question

The hare and the tortoise are at the starting line together. When the gun goes off, the hare moves off at a constant speed of 15 meters per second. (Ignore the acceleration required to get the animal to this speed.) The tortoise starts more slowly, but accelerates at the rate of 4 meters per second squared. How long will it be before the tortoise passes the hare?

thanks Swamy, i'll be shure to give you best

Answer 1

Let t be the time in seconds when the tortoise meets the hare.

15t = 0 + 1/2.4.t^2

since 15t is the distance covered by the hare and ut + 1/2 a.t^2 is the formula for the distance covered by the tortoise.

15t = 2t^2 or 2t = 15

t = 7 and a half seconds
(7.5 seconds)

Answer 2

One can obtain an unique answer to this question if one knows the initial speed of the tortoise. In the absence of that value, let us assume (to get a unique answer) that the initial speed of the tortoise is zero.

When they meet again, the two animals have covered the same distance in the same time "t".
So 15*t= (1/2)*4*t^2, where 4m/sec/sec is the acceleration of the tortoise.
or 15*t = 2*t^2. As t is not equal to 0, we can divide both sides by t to get
15=2*t or t = 7/2 sec = 3.5 sec.
So the distance traversed by each is 15*3.5 = 52.5 meter.

Hence the tortoise will pass the hare again after 3.5 seconds at a distance of 52.5 meters from the starting point.