#39; and distance '$x#39; is $t=\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. Tags and topics: Calculus for Physics,Differential Calculus,JEE MAIN,JEE Advanced,JEE Mains,KCET,Motion in a Straight Line,One Dimensional Motion - Non-uniform Acceleration,NCERT,NEET, JEE Main Phy."> #39; and distance '$x#39; is $t=\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. Tags and topics: Calculus for Physics,Differential Calculus,JEE MAIN,JEE Advanced,JEE Mains,KCET,Motion in a Straight Line,One Dimensional Motion - Non-uniform Acceleration,NCERT,NEET, JEE Main Phy."> #39; and distance '$x#39; is $t=\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. Tags and topics: Calculus for Physics,Differential Calculus,JEE MAIN,JEE Advanced,JEE Mains,KCET,Motion in a Straight Line,One Dimensional Motion - Non-uniform Acceleration,NCERT,NEET, JEE Main Phy.">
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Question

The relation between time '$t$' and distance '$x$' is $t=\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. The relation between acceleration $(a)$ and velocity $(v)$ is :

$a=-5 \alpha v^5$
$a=-3 \alpha v^2$
$a=-2 \alpha v^3$
$a=-4 \alpha v^4$

Solution

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