Answer
The distance is $4.0 \times 10^2 ~m$
Work Step by Step
We can use a force equation to find the acceleration:
$ma = mg ~sin(\theta)$
$a = g ~sin(\theta) = (9.80 ~m/s^2) \cdot ~sin(11^{\circ})$
$a = 1.9 ~m/s^2$
When a truck is heading up the ramp, the rate of deceleration is $1.9 ~m/s^2$. We can use kinematics to find the distance traveled by the truck:
$v_0 = (140 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s})$
$v_0 = 39~m/s$
$x = \frac{v^2-v_0^2}{2a} = \frac{0 - (39 ~m/s)^2}{(2)(-1.9 ~m/s^2)} = 4.0 \times 10^2 ~m$
The distance the truck travels up the ramp is $4.0 \times 10^2 ~m$
