JEE Main 202431 Jan, Morning shiftPhysicsMediumKinematics177× since 2002The relation between time 'ttt' and distance 'xxx' is t=αx2+βxt=\alpha x^2+\beta xt=αx2+βx, where α\alphaα and β\betaβ are constants. The relation between acceleration (a)(a)(a) and velocity (v)(v)(v) is :AOption A. a=−5αv5a=-5 \alpha v^5a=−5αv5BOption B. a=−3αv2a=-3 \alpha v^2a=−3αv2COption C. a=−2αv3a=-2 \alpha v^3a=−2αv3DOption D. a=−4αv4a=-4 \alpha v^4a=−4αv4CheckSkip · show solution
