Kinematics

JEE Main level, numerical answer, medium

13 of 308

Question 13 of 308

The displacement and the increase in the velocity of a moving particle in the time interval of tt to (t+1)s(t+1)s are 125m125m and 50ms−150m{s}^{-1}, respectively. The distance travelled by the particle in (t+2)ths(t+2{)}^{\text{th}}s is ___________ mm.

Show answer and solution

Answer: 175

Considering acceleration is constant, we can write
v=u+atv=u+at
As increase in velocity in 1s1s is given as 50ms−150m{s}^{-1}, therefore
⇒a=50ms−2\Rightarrow a=50m{s}^{-2}
Now, displacement covered between given time interval is,
125=ut+12at2125=ut+\dfrac{1}{2}a{t}^{2}
⇒125=u+a2\Rightarrow 125=u+\dfrac{a}{2}
⇒u=100ms−1\Rightarrow u=100m{s}^{-1}
Velocity after 1s1s will be, v=100+50=150ms−1v=100+50=150m{s}^{-1}
Therefore, displacement covered in (t+1)sto(t+2)s\left(t+1\right)s\text{to}\left(t+2\right)s will
∴Snth=vt′+12a(t′)2,wheret′=1s∴{S}_{{n}^{th}}=v{t}^{'}+\dfrac{1}{2}a{\left({t}^{'}\right)}^{2},\text{where}{t}^{'}=1s
=175m=175m

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Projectile motion simulationDrag the sliders and watch the numbers in this question change.Revise Kinematics & Projectile Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Kinematics (Rest and Motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.