A truck is initially moving at velocity $v$. The driver presses the brake in order to slow the truck to a stop. The brake applies a constant force $F$ to the truck. The truck rolls a distance $x$ before coming to a stop, and the time it takes to stop is $t$.

Which of the following expressions is equal the initial momentum of the truck (i.e. the momentum before the driver starts braking)?

$\textbf{(A) } Fx\\
\textbf{(B) }  Ft/2\\
\textbf{(C) } Fxt\\
\textbf{(D) } 2Ft\\
\textbf{(E) } 2Fx/v$
1. **Identify the initial momentum of the truck:**
   The initial momentum \( p \) of the truck is given by:
   \[
   p = mv
   \]
   where \( m \) is the mass of the truck and \( v \) is its initial velocity.

2. **Relate the force to the deceleration:**
   The force \( F \) applied by the brakes causes a deceleration \( a \). According to Newton's second law:
   \[
   F = ma
   \]
   Therefore, the deceleration \( a \) is:
   \[
   a = \frac{F}{m}
   \]

3. **Determine the stopping distance \( x \):**
   Using the kinematic equation for uniformly accelerated motion:
   \[
   v^2 = u^2 + 2ax
   \]
   where \( u \) is the initial velocity, \( v \) is the final velocity (which is 0 when the truck stops), and \( x \) is the stopping distance. Substituting \( v = 0 \) and \( a = -\frac{F}{m} \):
   \[
   0 = v^2 - 2 \left( \frac{F}{m} \right) x
   \]
   Solving for \( x \):
   \[
   x = \frac{mv^2}{2F}
   \]

4. **Determine the stopping time \( t \):**
   Using the kinematic equation:
   \[
   v = u + at
   \]
   where \( u = v \) and \( a = -\frac{F}{m} \):
   \[
   0 = v - \left( \frac{F}{m} \right) t
   \]
   Solving for \( t \):
   \[
   t = \frac{mv}{F}
   \]

5. **Relate the initial momentum to the given expressions:**
   We need to express the initial momentum \( mv \) in terms of the given quantities \( F \), \( x \), and \( t \).

   From the stopping time \( t \):
   \[
   mv = Ft
   \]

   From the stopping distance \( x \):
   \[
   x = \frac{mv^2}{2F}
   \]
   Solving for \( mv \):
   \[
   mv = \frac{2Fx}{v}
   \]

6. **Compare with the given options:**
   - Option (A) \( Fx \)
   - Option (B) \( \frac{Ft}{2} \)
   - Option (C) \( Fxt \)
   - Option (D) \( 2Ft \)
   - Option (E) \( \frac{2Fx}{v} \)

   The expression \( \frac{2Fx}{v} \) matches our derived expression for the initial momentum \( mv \).

The final answer is \(\boxed{\frac{2Fx}{v}}\).