Question

A 1.4 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 − (1.6 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?

Answer 1

Answer:

The value of c is 27.4 m/s²

Explanation:

Hi there!

Let´s write the position function:

x = 3.0 m + (4.0 m/s) · t + c · t² - (1.6 m/s³) · t³

The velocity of the particle is given by the derivative of the position function with respect to time:

dx/dt = v = 4.0 m/s + 2 · c · t - 4.8 m/s³ · t²

The acceleration of the particle is the derivative of the velocity function with respect to t:

dv/dt = a = 2 · c - 9.6 m/s³ · t

The applied force at t = 3.0 s is calculated as follows:

F = m · a

Where:

F = applied force.

m = mass of the particle.

a = acceleration.

Then:

F = m · a

36 N = 1.4 kg · a

36 N / 1.4 kg = a

a = 26 m/s²

We have derived the equation of the acceleration above:

a = 2 · c - 9.6 m/s³ · t

Then, using a = 26 m/s² and t = 3.0 s, we can solve the equation for c:

26 m/s² = 2 · c - 9.6 m/s³ · 3.0 s

26 m/s² + 9.6 m/s³ · 3.0 s  = 2 · c

54.8 m/s² = 2 · c

54.8 m/s² / 2 = c

c = 27.4 m/s²

The value of c is 27.4 m/s²