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Solution 8.6b

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Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get
Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get
 +
 +
<math>\mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+35\mathbf{i}+70\mathbf{j}=275\mathbf{i}-250\mathbf{j}</math>

Revision as of 18:36, 15 April 2010

We aassume the starting point is the origin. This means r0=0

The boat first accelerates to a point A say. We first must calculate the position of this point rA.

Using r=ut+21at 2+r0 with,

t=10, u=i+2j and from part a) a=0.5ij ms2

rA=(i+2j)10+21(0.5ij)10 2+0=35i+70j

The next stage has

t=40, a=0, r0=35i+70j, and u=6i8j as the final position and velocity of the first stage is the initial position and velocity of the second stage.

Using once again r=ut+21at 2+r0 we get

r=(6i8j)40+35i+70j=275i250j