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[In this question $i$ and $j$ are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin.] A ship $A$ moves with constant velocity $(3i-10j)\,\text{kmh}^{-1}$ . At time $t$ hours, the position vector of $A$ is $r\,\text{km}$ . At time $t=0$ , $A$ is at the point with position vector $(13i+5j)\,\text{km}$ . (a) Find $j$ $0$ in terms of $t$ .

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. [In this question

i

and

j

are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin.] A ship

A

moves with constant velocity

(3i-10j)\,\text{kmh}^{-1}

. At time

t

hours, the position vector of

A

is

r\,\text{km}

. At time

t=0

,

A

is at the point with position vector

(13i+5j)\,\text{km}

. (a) Find

j

0

in terms of

t

.

Initialize Equation: To find $r$ , we need to use the equation $r = r_0 + vt$ , where $r_0$ is the initial position vector and $v$ is the velocity vector.
Substitute Values: Given $r_0 = (13i + 5j) \, \text{km}$ and $v = (3i - 10j) \, \text{km/h}$ , we can plug these into the equation.
Distribute $t$ : So, $r = (13i + 5j) + t(3i - 10j)$ .
Combine Like Terms: Now we distribute $t$ across the velocity vector: $r = (13i + 5j) + (3ti - 10tj)$ .
Final Position Vector: Combine like terms to get the final position vector: $r = (13 + 3t)i + (5 - 10t)j$ .

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