The position particle function |RECTILINEARMOTIONJEEMain2019

JEE MAIN 2019

09-04-2019 S2

Question

The position of a particle as a function of time $t$, is given by $$ x(t)=a t+b t^2-c t^3 $$ where $a, b$ and $c$ are constants. When the particle attains zero acceleration, then its velocity will be :

Select the correct option:

$a+\frac{b^2}{4 c}$

$a+\frac{b^2}{3 c}$

$a+\frac{b^2}{2 c}$

$a+\frac{b^2}{c}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Hard

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