Question

At t=2 seconds, why is the acceleration of a position function x(t)=sin(2t^2 + 5t) equal to 129.56?

Answer 1

Acceleration = 129.56 (2 d.p.)

Explanation:

To find the acceleration of the position function x(t) = sin(2t² + 5t) at t = 2 seconds, we need to take the second derivative of the position function with respect to time (t) and then evaluate it at t = 2 seconds.

To find the first derivative of x(t) with respect to t, use the chain rule for differentiation.

`\boxed{\begin{array}{c}\underline{\text{Chain Rule for Differentiation}}\\\\\text{If\;\;$y=f(u)$\;\;and\;\;$u=g(x)$\;\;then:}\\\\\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}*\frac{\text{d}u}{\text{d}x}\end{array}}`

`\text{Let\;\;$x=\sin(u)$\;\;where\;\;$u=2t^2+5t$.}`

Differentiate the two parts separately:

`\frac{\text{d}x}{\text{d}u}=\cos(u)`

`\frac{\text{d}u}{\text{d}t}=4t+5`

Put everything back into the chain rule formula:

`x'(t)=\frac{\text{d}x}{\text{d}u}*\frac{\text{d}u}{\text{d}t}`

`x'(t)=\cos(u)*(4t+5)`

Substitute back in u = 2t² + 5t:

`x'(t)=\cos(2t^2+5t)*(4t+5)`

So, the first derivative, which represents the velocity function, is:

`\textbf{v}(t)=x'(t) = (4t + 5)\cos(2t^2 + 5t)`

To find the second derivative of x(t) with respect to t, use the product rule for differentiation.

`\boxed{\begin{array}{c}\underline{\sf Product\;Rule\;for\;Differentiation}\\\\\text{If}\;y=uv\;\text{then:}\\\\\frac{\text{d}y}{\text{d}x}=u\frac{\text{d}v}{\text{d}x}+v\frac{\text{d}u}{\text{d}x}\end{array}}`

`\textsf{Let}\;\;u=4t+5 \implies \frac{\text{d}u}{\text{d}t}=4`

`\textsf{Let}\;\;v=\cos(2t^2+5t) \implies \frac{\text{d}v}{\text{d}t}=-(4t+5)\sin(2t^2+5t)`

Put everything into the product rule formula:

`x''(t)=u\frac{\text{d}v}{\text{d}t}+v\frac{\text{d}u}{\text{d}t}`

`x''(t)=(4t+5)(-(4t+5)\sin(2t^2+5t))+4(\cos(2t^2+5t))`

`x''(t)=-(4t+5)^2\sin(2t^2+5t)+4\cos(2t^2+5t)`

So, the second derivative, which represents the acceleration function, is:

`\textbf{a}(t)=x''(t)=-(4t+5)^2\sin(2t^2+5t)+4\cos(2t^2+5t)`

Now, evaluate x''(t) at t = 2 seconds:

`x''(2)=-(4(2)+5)^2\sin(2(2)^2+5(2))+4\cos(2(2)^2+5(2))`

`x''(2)=-(4(2)+5)^2\sin(2(4)+5(2))+4\cos(2(4)+5(2))`

`x''(2)=-(8+5)^2\sin(8+10)+4\cos(8+10)`

`x''(2)=-(13)^2\sin(18)+4\cos(18)`

`x''(2)=-169\sin(18)+4\cos(18)`

`x''(2)=126.9168447...+2.64126683...`

`x''(2)=129.5581115...`

`x''(2)=129.56\; \sf (2\;d.p.)`

Therefore, the acceleration of a position function x(t) = sin(2t² + 5t) at t = 2 seconds is 129.56 (rounded to two decimal places).