Question

A particle is moving along x-axis such that its velocity v
varies with position as v = (2x + 1), where v is in m/s
and x is in m. The magnitude of acceleration of particle
at x = 1 m is​

Answer 1

Answer:

As shown in the given figure

equation of velocity and displacement is

V=−2x+4

acceleration a=V

dx

dv

=(−2x+4)(−2)

=4x−8

Answer 2

Answer:

The magnitude of the acceleration of the particle at x=1m is 6m/s^-2.

Explanation:

acceleration(a)=change in velocity(dv)/change in time(dt)

on multiplying and dividing numerator and denominator by dx we get

a=dv/dx x dx/dt

where dx is change in position

a=vdv/dx=v^2/2dx

given,

v=(2x+1)m/s and x=1 m

lets substitute this value in the equation

a=(2x+1)^2/2dx=(4x^2+4x+1)/2dx

a=1/2 x (4x^2+4x+1)/dx

a=1/2 x (8x+4)

a=1/2 x 12=6m/s^-2