Solution:
For a truck and a car moving with the same kinetic energy, the distance to stop under the same retarding force can be determined by using the equation:
\[
\text{Work done} = \Delta KE
\]
Since the initial kinetic energy is the same for both vehicles, the work done (force times distance) to bring them to rest will be equal. Thus, both vehicles come to rest in the same distance.
Statement II:
In Statement II, when a car changes its direction from east to north, its speed may remain constant, but its velocity is changing because velocity is a vector quantity. Since the direction of velocity is changing, the car has acceleration. Therefore, the acceleration is not zero.
\[
\Delta \vec{V} = \vec{V_f} - \vec{V_i}
\]
As velocity is changing, acceleration \( \vec{a} \neq 0 \).
Thus, Statement II is incorrect.
Thus, the correct answer is \( \boxed{1} \).
The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance covered by the object between \( t = 0 \) to \( t = 4s \)?
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: