JEE Main 2020MathematicsApplication of DerivativesEasyMCQ

JEE Main 2020 — Application of Derivatives Question with Solution

JEE Main 2020 (06 Sep Shift 1)

Question

The position of a moving car at time t is given by f(t)=at2+bt+c, t>0, where a, b and c are real numbers greater than 1. Then the average speed of the car over the time interval t1,t2 is attained at the point:

Choose an option

Show full solutionCorrect option: C
Correct answer
Ct1+t22

Step-by-step explanation

Given: position of the moving car at time t is =ft=at2+bt+c

So,

vavg=ft2-ft1t2-t1

⇒vavg=at22-t12+bt2-t1t2-t1

⇒vavg=at1+t2+b

The instantaneous speed is given by:

f't=2at+b

So, to get the point where average speed is equal to car's actual speed,

f't=vavg

⇒at1+t2+b=at+b

⇒t=t1+t22

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About this question

This is a previous-year question from JEE Main 2020, covering the Application of Derivatives chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.