Question

A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the time interval [0, 5], given that the initial velocity v(0) is 1/2 ft/sec. -4 1n(2)+4 1n(7)ft -ln(2)+ln(7)ft -31n(2) + 31n(7) ft -21n(2) + 21n(7) ft -1/2 ln(2) + 1/2ln(7)ft

Answer 1

`s(t)=(ln|7|+ln|2|)\,ft`

Explanation:

Acceleration is second derivative of distance and are related as:

`a(t)=(d^2s)/(dt^2)\\\\(d^2s)/(dt^2)=(-1)/((t+2)^2)\\`

Integrating both sides w.r.to t

`v(t)=(ds)/(dt)=(1)/(t+2) +C\\`

Using initial value

`v(0)=(1)/(2)\\\\(1)/(2)=(1)/(0+2) +C\\\\C=0\\\\(ds)/(dt)=(1)/(t+2)`

We have to calculate the distance covered in time interval [0,5], so:

`\int\limits^5_0 (ds)/(dt)=\int\limits^5_0 {(1)/(t+2)} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft`