Question

I'm not really sure how to go about creating the equation, can anyone help me?

A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the x axis be parallel to the ground and the positive y axis point upward. In the time interval from t=0 to t=4 s, the trajectory of the car along a certain section of the track is given by r(arrow on top) =A(1m/s)ti^+A[(1m/s3)t3−6(1m/s2)t2]j^, where A is a positive dimensionless constant.

Part B: Derive a general expression for the speed v of the car. Make sure that your expression would give the correct value for the speed (in m/s), but you don't need to put in anything explicitly about units (unlike what you see in the original expression for r(arrow on top)

Answer 1

The displacement vector (SI units) is

`\vec{r} =At\hat{i}+A[t^(3)-6t^(2)]\hat{j}`

The speed is a scalar quantity. Its magnitude is

`v= \sqrt{A^(2)t^(2)+A^(2)(t^(3)-6t^(2))^(2)} \\ v=A \sqrt{t^(2)+t^(6)-12t^(5)+36t^(4)} \\ v=At \sqrt{t^(4)-12t^(3)+36t^(2)+1}`

Answer: At√(t⁴ - 12t³ + 36t² + 1)