Question

Can someone help me out?

Answer 1

Q1:
`F(t) = m[6t \hat i + 12\hat j + 6t \hat k]`

Q2:

(a)
`a_T=4e^(2t)`

(b)
`a_N=2e^t`

Explanation:

Q1: What force is required so that a particle of mass m has the position function r(t) = t³i + 6t² j + t³k?

To determine the force required for a particle with the given position function, we need to find the particle's acceleration first and then apply Newton's second law of motion.

`a(t)=(dr^2(t))/(dt^2) = r''(t)`

Taking the first derivative of r(t), we have:

`r(t) = t^3 \hat i + 6t^2 \hat j + t^3 \hat k\\\\\\\\\Longrightarrow r'(t) = 3t^2 \hat i + 12t \hat j + 3t^2 \hat k`

Now taking the second derivative:

`r'(t) = 3t^2 \hat i + 12t \hat j + 3t^2 \hat k\\\\\\\\\therefore r''(t) = a(t) = 6t \hat i + 12\hat j + 6t \hat k`

Now we can apply Newtons second law:

`\rightarrow F(t)=ma(t)\\\\\\\\\therefore \boxed{ F(t) = m[6t \hat i + 12\hat j + 6t \hat k]}`

`\hrulefill`

Q2: Find the tangential and normal components of the acceleration vector, r(t) = ti + 2eᵗj+e²ᵗk.

To find the tangential (a) and normal components (b) of the acceleration vector, we can use the following formulas:

`a_T=(d\big|\big|v(t)\big|\big|)/(dt) \\\\\\a_N=√(\big|\big|a(t)\big|\big|^2-a_T^2)`

Part (a):

Finding v(t) by taking the derivative of r(t):

`r(t)= t\hat i + 2e^t \hat j + e^(2t) \hat k\\\\\\\\ \Longrightarrow r'(t)=v(t)= \hat i + 2e^t \hat j + 2e^(2t) \hat k\\\\\\\\\therefore v(t)= \hat i + 2e^t \hat j + 2e^(2t) \hat k`

Find the magnitude of the velocity function using the following formula:

`\big|\big|v(t)\big|\big|=√((v_x)^2+(v_y)^2+(v_z)^2) \\\\\\\\\Longrightarrow \big|\big|v(t)\big|\big|=√((1)^2+(2e^t)^2+(2e^2t)^2)\\\\\\\\\therefore \big|\big|v(t)\big|\big|= 2e^(2t)+1`

Now take the derivative of this function, we get:

`\Longrightarrow a_T=(d)/(dt)[2e^(2t)+1] \\\\\\\\\therefore \boxed{a_T=4e^(2t)}`

Part (b):

Start by finding a(t) by taking the derivative of v(t):

`v(t)= \hat i + 2e^t \hat j + 2e^(2t) \hat k\\\\\\\\\Longrightarrow v'(t)=a(t)= 2e^t \hat j + 4e^(2t) \hat k\\\\\\\\\therefore a(t)= 2e^t \hat j + 4e^(2t) \hat k\\`

Now calculate the magnitude of a(t):

`\Longrightarrow \big|\big|a(t)\big|\big|=\sqrt{(0)^2+(2e^t)^2+(4e^(2t))^2} \\\\\\\\\therefore \big|\big|a(t)\big|\big|=2e^t\sqrt{4e^(2t)+1}`

Substitute our expressions into the formula:

`\Longrightarrow a_N=\sqrt{(2e^t\sqrt{4e^(2t)+1})^2-(4e^(2t))^2}\\\\\\\\\therefore \boxed{a_N=2e^t}`