Question

Please help!! What do I do

Answer 1

a(5) = 1.080

a(10) = 0.393

a(20) = 0.122

Explanation:

Take the derivative of velocity v(t) -> acceleration a(t)

v(t) -> a(t)

Quotient Rule:
`(u*dv-du*v)/(u^2)`

`120*(x)/(5x+13)`

Make u= 5x+13 and v=x

`120*(x'*(5x+13)-x*(5x+13)')/((5x+13)^2)\\\\120*(1*(5x+13)-x*(5))/((5x+13)^2)\\\\120*((5x+13)-5x)/((5x+13)^2)\\120*(13)/((5x+13)^2) \\\\a(t)=(1560)/((5x+13)^2)`

Now Subsitute your a(5) , a(10) , a(20). Also, round to 3 decimal place.

a(5) = 1.08033241 = 1.080

a(10) = 0.3930461073 = 0.393

a(20) = 0.122170826 = 0.122

Answer 2

(a) 1.080 ft/s²

(b) 0.393 ft/s²

(c) 0.122 ft/s²

Explanation:

Acceleration is the derivative of velocity. You are being asked to differentiate the given function and evalutate the derivative at three different times. The function is a rational function, so the formula for the derivative of a ratio is applicable.

(u/v)' = (vu' -uv')/v²

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For the given velocity function, the accleration is ...

v(t) = (120t)/(5t +13)

a(t) = v'(t) = ((5t +13)(120) -(120t)(5))/(5t +13)² = 1560/(5t +13)²

(a)

At t=5, the acceleration is ...

a(5) = 1560/(5·5 +13)² = 1560/1444 ≈ 1.080 . . . ft/s²

(b)

At t=10, the acceleration is ...

a(10) = 1560/(5·10 +13)² = 1560/3969 ≈ 0.393 . . . ft/s²

(c)

At t=20, the acceleration is ...

a(20) = 1560/(5·20 +13)² = 1560/12729 ≈ 0.122 . . . ft/s²

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Additional comment

Many graphing calculators are capable of finding the numerical value of the derivative of a function.