# The relation between time t and distance x is t = αx^{2 }+ βx, where α & β are constants, then the retardation is

Sriyathmika asked

**Asnwer**

Given t = αx^{2} + βx

Differentiate both sides with respect to time (t)

dt/dt = 2αx.dx/dt + β.dx/dt

1 = 2αxv + βv

v.(2αx + β) = 1

(2αx + β) = 1/v

Again differentiate both side with respect to time (t)

2α.dx/dt = -v

2αv = v

acceleration = -2αv

dt/dt = 2αx.dx/dt + β.dx/dt

1 = 2αxv + βv

v.(2αx + β) = 1

(2αx + β) = 1/v

Again differentiate both side with respect to time (t)

2α.dx/dt = -v

^{-2}. dv/dt2αv = v

^{-2}. accelerationacceleration = -2αv

^{3}Hence, retardation= 2αv

^{3}
You must log in to post a comment.