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 kinetic energy of the truck (i.e. the kinetic energy before the driver starts braking)?

$\textbf{(A) } Fx\\
\textbf{(B) }  Fvt\\
\textbf{(C) } Fxt\\
\textbf{(D) } Ft\\
\textbf{(E) } \text{Both (a) and (b) are correct}$
1. **Identify the initial kinetic energy of the truck:**
   The initial kinetic energy \( K \) of the truck is given by the formula:
   \[
   K = \frac{1}{2}mv^2
   \]
   where \( m \) is the mass of the truck and \( v \) is its initial velocity.

2. **Determine the work done by the braking force:**
   The work \( W \) done by the braking force \( F \) over the distance \( x \) is given by:
   \[
   W = F \cdot x
   \]
   According to the work-energy theorem, the work done by the braking force is equal to the change in kinetic energy of the truck. Since the truck comes to a stop, its final kinetic energy is zero, and the work done by the brakes is equal to the initial kinetic energy of the truck.

3. **Relate the work done to the initial kinetic energy:**
   Since the truck comes to a stop, the work done by the braking force is equal to the initial kinetic energy:
   \[
   \frac{1}{2}mv^2 = F \cdot x
   \]
   Therefore, the initial kinetic energy of the truck is:
   \[
   K = Fx
   \]

4. **Evaluate the other options:**
   - Option (B) \( Fvt \) is incorrect because it does not correctly represent the work done by the braking force.
   - Option (C) \( Fxt \) is incorrect because it introduces an unnecessary factor of \( t \).
   - Option (D) \( Ft \) is incorrect because it does not correctly represent the work done by the braking force.
   - Option (E) is incorrect because only (A) is correct.

Conclusion:
The correct expression for the initial kinetic energy of the truck is \( Fx \).

The final answer is \( \boxed{Fx} \)