Physics · Motion In A Straight Line · NEET
Just differentiate v with respect to time: a = dv/dt. Example: if v = 3t^2 + 2t, then a = dv/dt = 6t + 2. At t = 2 s, a = 6(2) + 2 = 14 m/s^2. This is the most common case in NEET, so always check first whether v (or x) is written as a function of t.
Use a = v (dv/dx) when velocity is given in terms of position x (not time), such as v = k*sqrt(x) or v^2 = 4x. You cannot differentiate v by t directly because there is no t in the equation. The chain rule gives a = dv/dt = (dv/dx)(dx/dt) = v (dv/dx). Both formulas are correct; you pick the one that matches the variable you are given.
When t is written in terms of x, first get dt/dx = 2x + 1, so velocity v = dx/dt = 1/(2x+1). Then acceleration a = v (dv/dx). Here dv/dx = d/dx[(2x+1)^-1] = -2/(2x+1)^2. So a = [1/(2x+1)] * [-2/(2x+1)^2] = -2/(2x+1)^3. The minus sign means the particle is slowing down (retardation).
Twice. First differentiation gives velocity v = dx/dt, second gives acceleration a = dv/dt = d^2x/dt^2. Example: x = t^3 - 6t^2 gives v = 3t^2 - 12t and a = 6t - 12. Do not stop at velocity; acceleration is the second derivative of position with respect to time.
Distance is the integral of speed over time: distance = integral of v dt between the two times. For example, if v = At + Bt^2, distance from t = 1 to t = 2 is integral of (At + Bt^2) dt = [At^2/2 + Bt^3/3] from 1 to 2 = (3/2)A + (7/3)B. Differentiation gives acceleration; integration gives distance/displacement.
The relation between time t and position x of a particle moving along a straight line is given by t = x^2 + x. The acceleration of the particle is:
If the velocity of a particle is v = At + Bt^2, where A and B are constants, then the distance travelled by it between 1 s and 2 s is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Acceleration is the rate of change of velocity: a = dv/dt. Since v = dx/dt, you can also write a = d^2x/dt^2 (the second derivative of position with respect to time).
It is used when velocity depends on position x rather than time. Using the chain rule, a = dv/dt = (dv/dx)(dx/dt) = v (dv/dx). It is the standard tool for v = f(x) or t = f(x) problems.
No. a = dv/dt works for any motion, uniform or non-uniform. For uniform (constant) acceleration the derivative is a constant; for non-uniform acceleration it is a function of t or x.
Usually one direct question per year: differentiate a given v-t or x-t equation, or handle a t-x relation like t = x^2 + x (NEET 2025). Sometimes it is combined with integration to find distance (NEET 2016).
Next you study the three kinematic equations of uniformly accelerated motion (v = u + at, s = ut + 1/2 at^2, v^2 = u^2 + 2as), which are the constant-acceleration special case of these calculus relations.
