Physics · Motion In A Straight Line · NEET
To speed squared. From d = u^2 / 2a, if the retardation a stays the same, the stopping distance depends on u^2. So if you double the initial speed (u to 2u), the stopping distance becomes 4 times larger, not 2 times. This is why a fast car needs a much longer distance to stop. NEET loves this: they change the speed and ask for the new stopping distance, and the answer scales with the square.
In stopping problems you usually know the initial speed u, the final speed (0 at rest), and you want the distance s, but you do NOT know the time t. The equation v^2 = u^2 - 2as is the only kinematic equation with no time term, so it links u, v, a and s directly. Pick it whenever time is missing. If time is given or asked, use v = u + at or s = ut + (1/2)at^2 instead.
Retardation (also called deceleration) is acceleration that acts opposite to the motion, so it slows the body down. Uniform means it stays constant in size for the whole motion. If retardation is uniform, all three equations of motion apply, just with the acceleration written as a negative number (a = -a). Braking cars, bullets in wood, and objects sliding on a rough floor are usually treated as uniform retardation in NEET problems.
Set the final velocity to zero and rearrange. From v^2 = u^2 - 2as with v = 0, you get a = u^2 / 2s. So if a car moving at u = 20 m/s stops in s = 50 m, retardation a = (20)^2 / (2 x 50) = 400 / 100 = 4 m/s^2. Always keep units in SI (m/s and metres) before plugging in.
During braking the velocity vector points forward but the acceleration vector points backward (opposite to motion). Taking the direction of motion as positive, the acceleration comes out negative, which we call retardation. In the formula d = u^2 / 2a we usually plug in the MAGNITUDE of a (a positive number), because the minus sign is already handled when we wrote v^2 = u^2 - 2as with v = 0.
A bullet enters a wooden block with speed u and, after travelling 24 cm inside it, its speed reduces to u/3. Assuming uniform retardation, the further distance it travels before coming to rest is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
d = u^2 / 2a, where u is the initial speed and a is the magnitude of the uniform retardation. It comes from v^2 = u^2 - 2as with the final speed v = 0.
In the basic kinematic formula d = u^2 / 2a, mass does not appear, so if the retardation a is given, mass has no effect. Mass matters only when the retardation itself comes from a fixed braking force (a = F/m), which is a Laws of Motion extension, not a pure kinematics problem.
They mean the same thing: acceleration acting opposite to the direction of motion, which slows the body. We call its magnitude the retardation and write the vector as negative when the direction of motion is taken as positive.
Nine times longer. Since d is proportional to u^2, tripling u multiplies the stopping distance by 3^2 = 9 (assuming the same retardation).
Use v^2 = u^2 - 2as, because it is the only one of the three equations that has no time term. Stopping problems give speed and distance but usually not time.
