Question

In: Physics

Varible Force A 1.8 kg particle moves along an x axis, being propelled by a variable...

Varible Force A 1.8 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by x = 3.0 m + (4.0 m/s)t + ct² - (2.2 m/s³)t³, with x in meters and t in seconds. The factor c is a constant. At t = 3.0 s, the force on the particle has a magnitude of 36 N and is in the negative direction of the axis. What is c?
in m/s²

Expert Solution

The particle is moving along x-axis and mass of the particle is given

m = 1.8 kg

Position with respect to time

x = 3.0m + (4.0m/s)t +ct^{2 _} (2.2 m/s³)t³

The first derivative of position with respect to time is velocity and it's second deriative is acceleration.

So taking first derivative with respect to time ,

Taking second derivative ,

acceleration a =

At time t =3.0s , force is given by

F = -36N ( negative sign is because the force is acting in negative x axis opposite direction of the motin of the particle)

hence, ma = -36N

a = -36/m

a = -36/1.8

= -20m/s² ----------------(3)

we have acceleration interms of time from double derivative of position, so putting t =3 in equation 2 gives

a = 2c - 2.2m/s³*6*3 s

= 2c - 39.6m/s² ----------------(4)

equating (3) and (4) gives

2c -39.6m/s² = -20m/s²

2c =( -20+39.6)m/s²

2c = 19.6 m/s²

c = (19.6/2)m/s²

= 9.8 m/s²

Hence c is 9.8 m/s²

genius_generous answered 2 years ago

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