A=F/M Based problem - air track glider

[HJ^2]

At time t=0 s a 0.362-kg air track glider is moving rightward at 5.1 m/s. At t=2.5 s, it's going leftward at 5.4 m/s. Determine the magnitude and direction of the constant force that acted on the glider during this interval. Give your answer in Newtons. If the force is going rightward, your force should be positive. If the force is going leftward, your force should be negative.

There should be only one answer to this problem as it's being submitted on a website from one of many computer generated set problems.

I understand w=mg but not much past setting up the regular a=f/m equations rather than diving into Kinematic formulas to solve something Algebraically. In other words, I'm kind of clueless.

 

[gneill]

Hi HJ^2. Welcome to Physics Forums.

You'll have to pull out some formula other than w = mg. What other kinematics formulas have you studied? It looks like you've got an initial velocity, a final velocity, and a time interval to work with.

 

[HJ^2]

Hi HJ^2. Welcome to Physics Forums.

You'll have to pull out some formula other than w = mg. What other kinematics formulas have you studied? It looks like you've got an initial velocity, a final velocity, and a time interval to work with.

Sorry, this is my first time here.
Kinematic equations we have used include;
X_f=X_i+V_it + (at² / 2)
V_f=V_i+at
V_f²=V_i²+2a (X_f - X_i)

Recently, we've just covered A = F/M and f/n=M (coefficient of friction)

 

[gneill]

Xf=Xi+Vit + (at2 / 2)
Vf=Vi+at
Vf2=Vi2+2a (Xf - Xi)

So which of the above contains the variables for the quantities you've been given?

 

[HJ^2]

V_f=V_i+at ? So maybe plug those numbers in, solve for a, and then plug that into a=f/m?

 

[HJ^2]

Okay, I think I understand this! Thanks so much both of you