Question

Hi!A particle moves along a straight line, so its speed is () = ^2 − + 6, where t is the time measured in seconds and the speed is measured in meters timessecond.a) Calculate the distance traveled between the seconds t=1 and t=3

Answer 1

In this problem

the distance traveled between the seconds t=1 and t=3 is given by

`\int_1^3(t^2-t+6)dt=(50)/(3)\text{ m}`

The answer is

50/3 meters

or 16.67 meters

Explanation of integrals

In this problem we have

`\int_1^3(t^2-t+6)dt=\int_1^3t^2dt-\int_1^3tdt+\int_1^36dt`
`\begin{gathered} \int_1^3t^2dt=(t^3)/(3) \\ Evaluate\text{ at 3 and 1} \\ ((3)^3)/(3)-(1^3)/(3)=(27)/(3)-(1)/(3)=(26)/(3) \end{gathered}`
`\begin{gathered} -\int_1^3tdt=-(t^2)/(2) \\ evaluate\text{ at 3 and 1} \\ -(3^2)/(2)+(1^2)/(2)=-(9)/(2)+(1)/(2)=-4 \end{gathered}`
`\begin{gathered} \int_1^36dt=6t \\ evaluate\text{ at 3 and 1} \\ 6(3)-6(1)=12 \end{gathered}`

substitute

`\int_1^3t^2dt-\int_1^3tdt+\int_1^36dt=(26)/(3)-4+12=(50)/(3)`