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A particle is moving along a line such that when it is at the origin it has a velocity of 4m/s. If it begins to decelerate at the rate of a = (-1.5v

Therefore the givens are: v=4 m/s when t=0s when s=0m and v=? when t=2s when s=?

as the particle decelerates during this period by a factor of a =(-1.5v

**Problem #1.**A particle is moving along a line such that when it is at the origin it has a velocity of 4m/s. If it begins to decelerate at the rate of a = (-1.5v

^{1/2}) m/s^{2}, where v is in m/s, determine the particles position and celocity when t = 2s.Therefore the givens are: v=4 m/s when t=0s when s=0m and v=? when t=2s when s=?

as the particle decelerates during this period by a factor of a =(-1.5v

^{1/2}) m/s^{2}**All Available Equations are: a=(dv/dt), v=(ds/dt), ads=vdv. The other equations are for constant acceleration which doesn't apply here because it's changing.**

In my thinking, the only equation which would be available is a = (dv/dt) since you know "a" and can integrate "dv" to find the final velocity since you know both "t" values.

Therefore: (-1.5v

^{1/2})=dv/dt

2/3 = v

^{1/2}

v=0.816m/s

This value is wrong though because the answer in the book is 0.25m/s and s = 3.5m.

Ill be able to find the position once I get the velocity but I've thought about this too much for how simple it is and I've gotten nowhere.

Ball A is released from rest at a height of 40ft at the same time a second ball B is thrown upward starting at a height of 5ft from the ground. If the balls pass one another at a height of 20ft determine the speed at which ball B was thrown upward.

**Problem #2**Ball A is released from rest at a height of 40ft at the same time a second ball B is thrown upward starting at a height of 5ft from the ground. If the balls pass one another at a height of 20ft determine the speed at which ball B was thrown upward.

**All Available Equations are: a=(dv/dt), v=(ds/dt), ads=vdv. The other equations are for constant acceleration which can be applied in this case since a = g.**

Since I knew the initial velocity for ball A I used it to determine the time at which it would pass 20ft.

Using: S=S

_{0}+V

_{0}t + 1/2at

^{2}therefore

I changed the heights so that it was simpler by viewing it as if Ball be were at the origin (i.e. all distances are -5)

15ft=35ft+0t+1/2(+9.8)t

^{2},

I used +9.8 because the ball is falling therefore its helping to speed the ball up.

I got t = 2.02s.

I then used this answer to solve for the velocity of Ball B, the velocity it would have as it passed by 20ft. S=S

_{0}+ Vt + 1/2at

^{2}

V

_{0}=17.32

This velocity would be the velocity at t=2.02 when the ball is at 20ft not when it was thrown, therefore I used:

V

^{2}=V

_{0}

^{2}+2a(S-S

_{0})

V

_{0}2.45mft/s or 24.36ft/s depending on + or - gravity just

to check.

These are not correct because the answer is 31.4ft/s I think that my problem is in the last step that the ball is going up and coming down therefore it could pass 20ft with the other ball on the way down or the way up, depending, but I havn't been able to set it up to work out correctly.

I appreciate any help anyone could provide with either of these.