Given: t=αx2+βx
Differentiating w.r.t time, we get
dxdt=2αx+β
⇒v1=2αx+β
Differentiating w.r.t time again, we get
⇒−v21dtdv=2αdtdx
⇒dtdv=−2αv3[asv=dtdx]⇒a=dtdv=−2αv3
JEE Main 2024 — Physics Mechanics
The relation between time ‘t’ and distance ‘x’ is t=αx2+βx, where α and β are constants. The relation between acceleration (a) and velocity (v) is:
Held on 31 Jan 2024 · Verified 6 Jul 2026.
a=−2αv3
a=−5αv5
a=−3αv2
a=−4αv4
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