A particle travels along the x-axis with velocity v(t)=(2t)/(1+t^2) m/s for 0≤t≤3 seconds. If the particle’s position at t = 0 is x(0)=5, what is its position at t = 3?

7.3025850929

Step-by-step explanation:

v(t)=(2t)/(1+t^2)

To find the position we need to integrate the function

p(t) = ∫ v(t)

p(t) =∫(2t)/(1+t^2) dt

Using u substitution

u = 1+t^2

du =2t dt

p(t) =∫(du)/(u)

We know that the integral of 1/u du is ln |u|

p(t) = ln|u| +C

Substituting back for u

p(t) = ln|1 +t^2| +C

To find the value of C, we let t=0

p(0) = ln|1 +0| +C = 5

      = ln(1) +C =5

          0 +C =5     Therefore C=5

p(t) = ln|1 +t^2| +5

We want to find the position at t=3

p(3)  = ln|1 +3^2| +5

        = ln(10) +5

          =7.3025850929

ln(10) + 5

Step-by-step explanation:

x is the integral of v

Integral of 2t/(1 + t²) = ln(1+t²)

x(t) = ln(1+t²) + c

At t = 0, x = 5

5 = ln(1) + c

c = 5

x(t) = ln(1 + t²) + 5

At t = 3

x(3) = ln(1 + 3²) + 5

= ln(10) + 5

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