$(a)$ State Newton's first and third laws of motion.
$(b)$ $A$ car of mass $1800 \, kg$ moving with a speed of $10 \, m s^{-1}$ is brought to rest after covering a distance of $50 \, m$. Calculate the force acting on the car.

(D) First Law: It states, "Everybody in the universe stays in a state of rest or of uniform motion along a straight line unless compelled by an external force to change its state."
Third Law: It states, "To every action, there is an equal (in magnitude) and opposite (in direction) reaction." In other words, action and reaction are equal and opposite.
$(b)$ Given: Mass $m = 1800 \, kg$, initial velocity $u = 10 \, m s^{-1}$, final velocity $v = 0$, and distance $S = 50 \, m$.
Using the third equation of motion: $v^2 - u^2 = 2aS$
$0^2 - (10)^2 = 2 \times a \times 50$
$-100 = 100a$
$a = -1 \, m s^{-2}$ (The negative sign indicates retardation).
Now, calculating the force using Newton's second law: $F = ma$
$F = 1800 \times (-1)$
$F = -1800 \, N$
The force acting on the car is $1800 \, N$ in the direction opposite to the motion.