Physics · Motion In A Straight Line · NEET
On a v-t graph of uniform acceleration, the line is straight. Acceleration = slope = (change in velocity)/(time). So a = (v - u)/t. Rearrange: v - u = at, giving v = u + at. Here u is the initial velocity (the y-intercept) and v is the final velocity after time t. The whole first equation is just 'slope of the v-t line'.
For uniform acceleration the v-t graph is a straight line from height u to height v. The area under it (which equals displacement) is a rectangle of height u plus a triangle on top. Rectangle area = u × t. Triangle area = ½ × base × height = ½ × t × (v - u) = ½ × t × at = ½at². Add them: s = ut + ½at². So displacement = area under the graph, split into rectangle + triangle.
Two ways. Graphical: the area can also be written as average velocity × time = ((u+v)/2)·t. Put t = (v - u)/a from the first equation: s = ((u+v)/2)·((v-u)/a) = (v² - u²)/(2a), which rearranges to v² = u² + 2as. Calculus: write a = v(dv/ds), so v dv = a ds, and integrate both sides from u to v and 0 to s. This equation is useful because it has no t in it — use it when time is neither given nor asked.
Start from a = dv/dt. Integrate: ∫dv = ∫a dt, and since a is constant, v - u = at, so v = u + at. Next use v = dx/dt, so dx = v dt = (u + at)dt. Integrate: x - x₀ = ut + ½at². Finally write a = v(dv/dx), so v dv = a dx, integrate to get (v² - u²)/2 = a(x - x₀), i.e. v² = u² + 2as. The calculus method also works for variable acceleration if a is a function you can integrate — the graphical shortcut does not.
No. The graphical derivation assumes the v-t graph is a straight line, and the calculus one pulls 'a' out of the integral as a constant. Both need acceleration to be constant. If acceleration changes with time or position, you must integrate the actual a(t) or a(x) directly (calculus method), not plug into v = u + at.
If the velocity of a particle is v = At + Bt², where A and B are constants, then the distance travelled by it between 1 s and 2 s is:
The relation between time t and position x of a particle moving along a straight line is given by t = x² + x. The acceleration of the particle is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Learn both. The graphical method is fast and intuitive for the standard constant-acceleration equations and for graph-reading questions. The calculus method is essential when acceleration is not constant (v or a given as a function of t or x), which NEET tests almost every year.
u = initial velocity, v = final velocity, a = acceleration (constant), s (or x) = displacement, t = time. NCERT writes u as v₀. All five quantities are linked by the three equations; each equation is missing exactly one of them.
For a tiny time dt, displacement is v·dt, which is a thin strip of area under the v-t curve. Adding all strips (integrating) gives the total area = total displacement. This is the core idea behind the graphical derivation of s = ut + ½at².
Yes. Free fall is constant acceleration a = g (about 9.8 or 10 m/s² downward), so all three equations apply — just put a = g (or -g) with a chosen sign convention. Many NEET free-fall problems are direct applications of these derived equations.
There are three main ones. v = u + at has no s. s = ut + ½at² has no v. v² = u² + 2as has no t. Choosing the right one means picking the equation that skips the quantity you neither know nor need.
