Question

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters, t in seconds, and c and b are positive constants. What are the units of (a) constant c and (b) constant b? Find a formula in terms of c, b, and t of the (c) velocity v and (d) acceleration a. (e) At what time t ≥ 0 does the particle reach its maximum x value?

Answer 1

The units of constant c are in meters per second squared per second squared (m/s^2). The units of constant b are in meters per second cubed (m/s^3). The velocity can be found by taking the derivative of the position function, and the acceleration can be found by taking the derivative of the velocity function. To find the time when the particle reaches its maximum x value, set the velocity equal to zero and solve for t.

Step-by-step explanation:

(a) The units of constant c can be determined by analyzing the equation x = ct2 - bt3. Since x is in meters and t is in seconds, the units of x/t2 will be in meters per second squared. Therefore, the units of constant c will be in meters per second squared per second squared (m/s2).

(b) Similar to part (a), the units of constant b can be determined by analyzing the equation x = ct2 - bt3. Since x is in meters and t is in seconds, the units of x/t3 will be in meters per second cubed. Therefore, the units of constant b will be in meters per second cubed (m/s3).

(c) The velocity v can be found by taking the derivative of the position function x with respect to time t. In this case, v = d/dt(ct2 - bt3) = 2ct - 3bt2.

(d) The acceleration a can be found by taking the derivative of the velocity function v with respect to time t. In this case, a = d/dt(2ct - 3bt2) = 2c - 6bt.

(e) To find the time when the particle reaches its maximum x value, we can set the velocity v equal to zero and solve for t. 2ct - 3bt2 = 0. Solving this equation will give us the time t when the particle reaches its maximum x value.

Answer 2

(a):
`\rm meter/ second^2.`

(b):
`\rm meter/ second^3.`

(c):
`\rm 2ct-3bt^2.`

(d):
`\rm 2c-6bt.`

(e):
`\rm t=(2c)/(3b).`

Step-by-step explanation:

Given, the position of the particle along the x axis is

`\rm x=ct^2-bt^3.`

The units of terms
`\rm ct^2` and
`\rm bt^3` should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of
`\rm ct^2=meter`

Therefore, unit of
`\rm c= meter/ second^2.`

(b):

Unit of
`\rm bt^3=meter`

Therefore, unit of
`\rm b= meter/ second^3.`

(c):

The velocity v and the position x of a particle are related as

`\rm v=(dx)/(dt)\\=(d)/(dx)(ct^2-bt^3)\\=2ct-3bt^2.`

(d):

The acceleration a and the velocity v of the particle is related as

`\rm a = (dv)/(dt)\\=(d)/(dt)(2ct-3bt^2)\\=2c-6bt.`

(e):

The particle attains maximum x at, let's say,
`\rm t_o`, when the following two conditions are fulfilled:

1. 
   `\rm \left ((dx)/(dt)\right )_(t=t_o)=0.`
2. 
   `\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.`

Applying both these conditions,

`\rm \left ( (dx)/(dt)\right )_(t=t_o)=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = (2c)/(3b).`

For
`\rm t_o = 0`,

`\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6\cdot 0=2c`

Since, c is a positive constant therefore, for
`\rm t_o = 0`,

`\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)>0`

Thus, particle does not reach its maximum value at
`\rm t = 0\ s.`

For
`\rm t_o = (2c)/(3b)`,

`\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6b\cdot (2c)/(3b)=2c-4c=-2c.`

Here,

`\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.`

Thus, the particle reach its maximum x value at time
`\rm t_o = (2c)/(3b).`