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The relationship between the distance x and time t is t =ax^2+bx where a and b are constants.Find the acceleration of the particle.

The relationship between the distance x and time t is t =ax^2+bx where a and b are constants.Find the acceleration of the particle.

Grade:11

2 Answers

Harsh jain
16 Points
6 years ago
ACC. is d2x /dt2On diff. The eq. 1 We have 1 = 2a×v + bOn further diff.0 = 2a× acc. + 0We have acc. =0
Neelabh Jaiswal
11 Points
6 years ago
Given t=ax^2+bxOn diff the above equationdt/dx=2ax+b.............1 v^-1=2ax+b...............3Further diff we get-v^-2dv=2adxOn simplificationdv/dxv^2=-2aOn cross multiplicationdv/dx=-2av^2......... equation 2Since dx=dt/2ax+b from 1Put it in equation 2dv(2ax+b)/dt=-2av^2Since 2ax+b=v^-1.... from 3dv/dtv=-2av^2 since dv/dt =a(acceleration)a/v=-2av^2a=-2av^3

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