Finding Velocity by Differentiating a Position-Time Equation (v = dx/dt)
Physics · Motion In A Straight Line · NEET
When position x is given as an equation in time t (like x = at + bt²), the velocity at any instant is the derivative: v = dx/dt. You differentiate the position equation once with respect to t, then put in the value of t to get the velocity at that moment. Memory hook: "Position tells you WHERE, one derivative tells you HOW FAST."
On a position-time (x-t) graph, the instantaneous velocity at any point P equals the slope of the tangent line there. Differentiating the position equation, v = dx/dt, computes exactly this slope at every instant.
Your doubts, answered
How do I find velocity if only the position equation x(t) is given?
Differentiate the position equation once with respect to time: v = dx/dt. Example: if x = 3t² + 2t, then v = dx/dt = 6t + 2. This v is the instantaneous velocity, valid at every instant t. To get velocity at a particular moment, substitute that value of t. At t = 2 s: v = 6(2) + 2 = 14 m/s.
Do I use the power rule term by term?
Yes. Differentiate each term separately using d/dt(tⁿ) = n·tⁿ⁻¹. A constant term (like +5) differentiates to 0 because it does not change with time. For x = 5t³ − 4t + 7, you get v = 15t² − 4 + 0 = 15t² − 4. The constant 7 simply disappears in velocity.
What is the difference between finding v(t) and finding v at one instant?
Differentiating gives you v as a full equation, v(t), which works for ALL times. That is the instantaneous velocity function. If the question asks for velocity 'at t = 3 s', do the differentiation first, THEN substitute t = 3. Never substitute the number before you differentiate, or you lose the t and cannot take the derivative.
Do I differentiate or integrate here?
Going DOWN the ladder (position to velocity to acceleration) you DIFFERENTIATE. Going UP (acceleration to velocity to position) you INTEGRATE. So position to velocity is differentiation: v = dx/dt. Only if you are given velocity and asked for distance/displacement do you integrate.
Two particles have position equations. How do I find when their velocities are equal?
Differentiate each position equation to get v₁(t) and v₂(t), then set v₁ = v₂ and solve for t. This is a real NEET 2016 question. For x_P = at + bt² and x_Q = ft − t², you get v_P = a + 2bt and v_Q = f − 2t; equating gives t = (f − a) / 2(1 + b).
⚠️ The NEET trap ✗ For x = 2t³ − 5t at t = 2 s, students plug in first: x = 2(8) − 10 = 6, then treat 6 as velocity. This is wrong — 6 is a position, not a velocity. ✓ Differentiate first: v = dx/dt = 6t² − 5. THEN substitute t = 2: v = 6(4) − 5 = 19 m/s. Always differentiate, then substitute. 🧠 Substituting the time value BEFORE differentiating.
Real NEET questions
2016
Two cars P and Q start from a point at the same time in a straight line and their positions are represented by x_P = at + bt² and x_Q = ft − t². At what time do the cars have the same velocity?
A · (f − a) / 2(1 + b) ✓
B · (a + f) / 2(b − 1)
C · (a − f) / (1 + b)
D · (a + f) / 2(1 + b)
Solution: Find each velocity by differentiating position. Car P: v_P = dx_P/dt = a + 2bt. Car Q: v_Q = dx_Q/dt = f − 2t. Set the velocities equal: a + 2bt = f − 2t. Group the t terms: 2bt + 2t = f − a, so t(2b + 2) = f − a. Therefore t = (f − a) / 2(1 + b). Answer: A.
2016
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:
A · (3/2)A + (7/3)B ✓
B · 3A + 7B
C · A/2 + B/3
D · (3/2)A + 4B
Solution: Here velocity is given, so to get distance you INTEGRATE (the reverse of differentiation): distance = ∫v dt from 1 to 2. Integrate: ∫(At + Bt²) dt = At²/2 + Bt³/3. Evaluate from 1 to 2: (A/2)(2² − 1²) + (B/3)(2³ − 1³) = (A/2)(4 − 1) + (B/3)(8 − 1) = (3A/2) + (7B/3). Answer: A. This is the mirror skill — position to velocity is differentiate, velocity to distance is integrate.
2026
A particle moves along a straight line with position s(t) = αt² − βt + γ, where α = 1 m/s², β = 6 m/s, γ = 5 m. The average speed of the particle from t = 0 to t = 6 s is:
A · 12
B · 6
C · 3 ✓
D · 0
Solution: First differentiate position to get velocity: v = ds/dt = 2αt − β = 2t − 6. Velocity is zero when 2t − 6 = 0, i.e. t = 3 s — the particle reverses direction here, so distance ≠ displacement. Split the trip. On [0, 3] and [3, 6] the speed magnitude reaches 6 at the ends (at t = 0 and t = 6, |v| = 6). Total distance = area of two v-t triangles = ½(3)(6) + ½(3)(6) = 9 + 9 = 18 m. Average speed = distance / time = 18 / 6 = 3 m/s. Answer: C. Trap: average VELOCITY here is 0 (option D) because s returns near start — but the question asks average SPEED.
Solved Motion In A Straight Line NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula to find velocity from a position-time equation?
v = dx/dt. You differentiate the position equation x(t) once with respect to time. The result v(t) is the instantaneous velocity at any instant.
Is velocity the first derivative or second derivative of position?
Velocity is the FIRST derivative of position: v = dx/dt. Acceleration is the second derivative of position (or first derivative of velocity): a = dv/dt = d²x/dt².
What happens to a constant term when I differentiate position?
A constant term becomes zero. For x = 4t² + 9, the +9 does not change with time, so v = dx/dt = 8t + 0 = 8t. Constants only shift the starting position, not the velocity.
How do I find velocity at a specific instant like t = 4 s?
Differentiate the position equation first to get v(t), and only THEN substitute t = 4. Never substitute the number before differentiating, or you cannot take the derivative.
Why does NEET keep asking this instead of just using v = u + at?
v = u + at only works for uniform (constant) acceleration. When position is a cubic or higher (like x = t³), acceleration changes with time, so you MUST use calculus v = dx/dt. NEET tests this to check if you know when the standard equations fail.
