Force on 500g Particle: Velocity at t = 2s

[bestchemist]

A 500g particle has velocity vx = -4.0m/s at t = - 2 s. Force Fx = (4−(t/s)2) N is exerted on the particle between t = - 2 s and t = 2 s. This force increases from 0 N at t = - 2 s to 4 N at t = 0 s and then back to 0 N at t = 2 s. What is the particle’s velocity at t = 2 s?
I'm not sure if the answer is 4 m/s. can someone check if I do it right?

V² = V₀² + at
v= v₀+ (F/m)t
v = -4m/s + (0/0.5)*2
v= 4m/s

 

[mfb]

v= v₀+ (F/m)t requires a constant force, or you have to calculate the average force. You cannot use the force at some arbitrary point in time here.

The first equation looks odd.
How did you get the last line from the line before?

 

[bestchemist]

It doesn't have constant force so which formula do I use?

v = 4 m/s?
I got that from plug in v₀²
v₀ = -4 m/s
so I thought it would be 4 m/s since I have to take square root to get v
Is it wrong? lol

 

[mfb]

It doesn't have constant force so which formula do I use?

How are the velocity as function of time and the force related?

Alternatively, calculate the average force.

I got that from plug in v₀²
v₀ = -4 m/s

The first formula was wrong anyway. You can see it if you check the units.

so I thought it would be 4 m/s since I have to take square root to get v
Is it wrong? lol

v²=16m²/s² has two solutions: v=4m/s and v=-4m/s. But the formula you used was not right anyway.

 

[bestchemist]

I figure it out! Thank you :)