4. A young girl gives her toboggan a push of 4.0 m/s up a hill. It slides up the hill slowing down at an acceleration of 8.0 m/s- [down). It comes to a stop and then slides back towards

her speeding up at the same rate as it slowed down on the way up. If the girl has to run 48 m
down the hill from where it first was pushed to get to where her sled stopped, find the
elapsed time for the journey.

Rereading the question, I think we have all the information we need.

At the starting position, the girl pushes the sled up the hill with speed 4.0 m/s and it slows to a stop uniformly with acceleration 8.0 m/s², then starts coming back down with acceleration 8.0 m/s², and comes to another stop at the bottom of the hill 48 m away from the starting position. If we take up-the-hill to be the positive direction, then we're saying the final position of the sled was -48 m from the origin. The acceleration throughout will have a negative sign.

After being pushed, the sled has a velocity v at time t of

v = 4.0 m/s + (- 8.0 m/s²) t

It comes to a stop when v = 0 :

0 = 4.0 m/s + (- 8.0 m/s²) t

(8.0 m/s²) t = 4.0 m/s

t = (4.0 m/s) / (8.0 m/s²)

t = 0.50 s

In this time, it undergoes a displacement x of

x = (4.0 m/s) t + 1/2 (- 8.0 m/s²) t²

so that, moving up the hill, it covers a distance of

x = (4.0 m/s) (0.50 s) + 1/2 (- 8.0 m/s²) (0.50 s)²

x = 1.0 m

Using the same origin, when the sled begins to slide back down the hill, its displacement from the origin is given by

x = 1.0 m + 1/2 (- 8.0 m/s²) t²

It slides 48 m down the hill from the origin, i.e. - 48 m from the origin, so that

- 48 m = 1.0 m + 1/2 (- 8.0 m/s²) t²

- 49 m = (- 4.0 m/s²) t²

t² = (49 m) / (4.0 m/s²)

t = 3.5 s

So, the sled travels for a total of 4.0 s up then down the hill.

because if the tool that you are using is not balanced it there would be false mass.

i need with 1.3.2 investigate cycling o2 and co2 apex

we need to see it