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Friday, 9 October

JEE Main 2020 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 8 January 2020, Shift 1, Physics Q21

Question 21 of 75 in this shift, Physics question 21 of 25, Section B.

A particle is moving along the xx-axis with its coordinate with time 't' given by x(t)=10+8t−3t2x(t) = 10+8t-3t^2. Another particle is moving along the yy-axis with its coordinate as a function of time given by y(t)=5−8t3y(t) = 5-8t^3. At t=1t = 1 s, the speed of the second particle as measured in the frame of the first particle is given as v\sqrt{v}. Then vv (in m/s) is ______.

Official answer

580.00

NTA final key (Jan 2020).

Same topic in other shifts

All Kinematics in 2D questions
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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.