MixedJEE Physics · Original learning card10 original chapter questions

Velocity-Time Graphs

For v(t), acceleration is dv/dt and displacement is the signed integral integral(v dt). Distance is integral(|v| dt), so negative graph regions contribute positively to distance.

Why this shows up in the exam

Piecewise acceleration · Distance versus displacement comparisons · Catch-up problems represented graphically

Learn the idea

A velocity-time graph gives acceleration by slope and displacement by signed area. The graph's height says the current signed velocity. The area accumulated above the time axis moves position forward, while area below it moves position backward.

🧠 Memory hook: On v-t, slope is a and signed area is displacement.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Delta x = integral(v dt) — signed area under a velocity-time curve
  • distance = integral(|v| dt) — sum of absolute areas
  • a = dv/dt — slope of the velocity-time graph

How to approach it

  1. 1Split the graph at corners and axis crossings
  2. 2Compute signed geometric areas
  3. 3Take absolute areas only for total distance

Common slip-ups that cost marks

  • •Adding negative area as negative distance
  • •Reading area as acceleration
  • •Ignoring time-scale or velocity-scale units

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 10

A particle has initial speed 2 m/s and constant acceleration 3 m/s^2 for 4 s. What distance does it cover?

Take a timed JEE Physics sectional mock