Physics · Motion In A Straight Line · NEET
Use the third equation v² = u² + 2as. It links final velocity, initial velocity, acceleration and displacement with no time in it. So when a NEET question gives you speeds and distance but no time (like a bullet stopping in a block, or a ball hitting the ground), reach for v² = u² + 2as first. If time IS given and you need displacement, use s = ut + (1/2)at².
s is displacement, not distance. The three kinematic equations are vector equations along one line, so s is the net change in position. It can be zero or negative. Distance (total path length) equals s only when the body never reverses direction. This is exactly why average velocity (uses displacement s) and average speed (uses total distance) can differ. For a ball thrown up that comes back, s over the full flight is 0 but the distance is not.
g is not fixed as plus or minus; you fix a sign convention first, then g follows it. If you take the upward direction as positive, acceleration = -g (it points down), so a = -9.8 m/s². For an object simply dropped, it is cleaner to take downward as positive, then a = +g and you write v² = u² + 2gh directly. Pick one convention per problem and use it for u, v, a and s together.
Distance in n seconds is the total covered from t = 0 to t = n, given by s = un + (1/2)an². Distance in the nth second is only the slice during that one second (from t = n-1 to t = n), given by s_n = u + (a/2)(2n-1). Notice s_n has units of metres even though it looks like a velocity formula, because it is distance per that one second. For a body starting from rest these slices go in the ratio 1 : 3 : 5 : 7.
No. All three kinematic equations assume acceleration is constant (uniform). If a changes with time or position, you must use calculus: v = dx/dt, a = dv/dt, and a = v(dv/dx). NEET tests this directly when it gives x = f(t) or v = f(t) or a t–x relation. In those cases differentiate or integrate; do not plug into v = u + at.
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground with a velocity of 80 m/s. The height of the tower is (g = 10 m/s²):
A vehicle travels half the distance with speed v and the remaining half distance with speed 2v. Its average speed is:
A block slides from rest down a smooth incline at t = 0. If S_n is the distance travelled from t = (n−1) s to t = n s, the ratio S_n / S_(n+1) is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
v = u + at, s = ut + (1/2)at², and v² = u² + 2as. Here u is initial velocity, v is final velocity, a is constant acceleration, t is time and s is displacement. They are valid only for uniform (constant) acceleration.
s_n = u + (a/2)(2n − 1). It gives the distance covered only during the nth second, not the total up to n seconds. For a body starting from rest the values follow the ratio 1 : 3 : 5 : 7.
Just replace a with g in the three equations. Common forms are v = gt, h = (1/2)gt², and v² = 2gh for a body dropped from rest. If it is thrown, keep u and mind the sign of g using your chosen positive direction.
Average speed = total distance / total time (a scalar, always positive if the body moves). Average velocity = total displacement / total time (a vector, can be zero or negative). They are equal only when the motion is along one direction without reversing.
When acceleration is not constant. If x, v or a is given as a changing function of time or position, use calculus: v = dx/dt, a = dv/dt or a = v(dv/dx). NEET regularly sets such questions to catch students who blindly apply v = u + at.
