{ The subscripts are in curly brackets }

A mass m is at rest on a horizontal frictionless surface at t=0. Then a constant force F{0} acts on it for a time t{0}. Suddenly, the force doubles to 2F{0} and remains constant until t=2t{0}. Determine the total distance D traveled from t=0 to t=2t{0}. Express your answer in terms of F{0}, t{0}, and m.

Get the acceleration from F = ma

During the first period,
a = F{0}/m and the displacement is
(1/2)*(F{0}/m)*t{0}^2

During the second period it travels twice as far, since the period of acceleration at the higher (doubled) rate is the same.

Add the two displacements

To determine the total distance traveled from t=0 to t=2t{0}, we need to break down the motion into two parts: the initial motion when the force is F{0} and the subsequent motion when the force is 2F{0}.

First, let's consider the initial motion.

When the force F{0} acts on the mass, it will accelerate according to Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, since the force is constant, the acceleration will also be constant.

Using the equation of motion, s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity (which is zero in this case since the mass is at rest), t is the time, and a is the acceleration, we can find the displacement during this period.

Since the mass is at rest initially, the initial velocity u is zero. So the equation simplifies to s = (1/2)at^2.

The time t spent during the initial period is t{0}, and the acceleration a can be found using F = ma, where F = F{0}.

Therefore, a = F{0}/m.

Plugging in these values into the equation for displacement, s = (1/2)(F{0}/m)t{0}^2.

Next, let's consider the subsequent motion when the force doubles to 2F{0}.

Again using the equation of motion, s = ut + (1/2)at^2, where u is the initial velocity, t is the time, and a is the acceleration, we can find the displacement during this period.

Since the force is double, the acceleration a will also be double. Therefore, a = 2F{0}/m.

The time t spent during this subsequent period is t{0}, as given.

Now, since the mass was initially at rest, the initial velocity u for the subsequent motion can be found using the equation v = u + at, where v is the final velocity at t = 2t{0}. Since v = 0 (since the mass comes to rest at t = 2t{0}), we have u + (2F{0}/m)t{0} = 0, from which u = -(2F{0}/mt{0}).

Plugging in these values into the equation for displacement, s = (-(2F{0}/mt{0}))t{0} + (1/2)(2F{0}/m)t{0}^2.

Now, to find the total displacement, we simply add the displacement during the initial motion and the displacement during the subsequent motion.

Total displacement, D = s (initial) + s (subsequent)
= (1/2)(F{0}/m)t{0}^2 + (-(2F{0}/mt{0}))t{0} + (1/2)(2F{0}/m)t{0}^2

Simplifying the expression, we get:

D = (1/2)(F{0}/m)t{0}^2 - (2F{0}/m)t{0}^2 + (F{0}/m)t{0}^2

= (1/2 - 2 + 1)(F{0}/m)t{0}^2

= (-3/2)(F{0}/m)t{0}^2

Therefore, the total distance traveled from t=0 to t=2t{0} is (-3/2)(F{0}/m)t{0}^2.