Physics Questions, Explain please?

Question

A jogger with a constant velocity of 4.0 meters/second runs by a stationary dog. After 1 second, the dog decides to chase the jogger. The dog accelerate at 1.5m/s².

(a)How long does it take the dog to catch the jogger?
(b) How far away from the spot where the dog was sitting has the jogger gone when she is caught by the dog?
Note: assume the jogger has a constant velocity.

Answer 1

let time be (t=0) when they cross each other
let they both meet at (t=t)
-------------------------
distance travelled by jogger = s = v*t = 4t
distance covered by dog = s1 = u (t-1) + 0.5 f(t - 1)^2
s1 = 0+ 0.5 * 1.5 (t - 1)^2 = 0.75 (t - 1)^2
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dog will move for (t-1) sec and jogger (t sec)
-------------------------
when they meet s = s1
4 t = 0.75 (t - 1)^2 = 3 (t - 1)^2 /4
3 (t - 1)^2 - 16 t =0
3[t^2+1-2t] -16 t=0
3t^2- 22 t +3]=0 >>> solve for t (+ve)
t = [22+- sqrt(484 - 36)]/6 = [22+- 21.17]/6
t1 = 0.138 >>> neglect because t is not even 1 sec
t2 = 7.195 s >>>>>>
EDITED>>>>>>>>>>
dog took time = (t-1) = 7.195 - 1 = 6.195 sec
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jogger travelled s = v t2 = 4*7.195 = 28.78 meter
dog travelled s1 = 0.75*(6.195)^2 = 28.78 meter
same

Answer 2

Let t be the time when the dog catches the jogger.

s = ut + 1/2 at^2 is the distance that the dog travelled.

= 4 + 4t (since the jogger is maintaining a constant velocity and the dog starts 1 second after the jogger passed by)

u = 0.

s = 1/2 x 1.5 x t^2 = 4 + 4t

or 0.75t^2 - 4t - 4 = 0

This is a quadratic equation in t and has two roots t1 and t2.

t1 = [4 + sqrt(14 + 12)] / 1.5

= (4 + 5.1) / 1.5 = 9.1 / 1.5 = 6.066 seconds

The dog catches up with (or catches if you prefer) the jogger 6.066 seconds, after it starts chasing.

t2 can be ignored as having no physical significance.

Distance from the original spot = 4 + 24.26 = 28.26 m