Question

the displacement of a boat relative to water is represented by (3t)i+4(t2-1)j(read as 4 (t square-1)j) and that of water relative to ground is i-(et)j (read as i-(e power t)j).what is the velocity of the boat relative to ground if i&j represent 1km/hour east and north respectively?

Answer 1

the displacement of a boat relative to water is represented by (3t)i+4(t2-1)j(read as 4 (t square-1)j) and that of water relative to ground is i-(et)j (read as i-(e power t)j).what is the velocity of the boat relative to ground if i&j represent 1km/hour east and north respectively?

Solution

The displacement of a boat relative to water $\overrightarrow{d_1}$ is represented by

$$\overrightarrow{d_1} = (3t)i + 4(t^2 - 1)j.$$

The displacement of water relative to ground $\overrightarrow{d_2}$ is represented by

$$\overrightarrow{d_2} = i - (e^t)j.$$

The displacement of a boat relative to ground $\overrightarrow{d_3}$ is the sum of displacements of a boat relative to water $\overrightarrow{d_1}$ and that of water relative to ground $\overrightarrow{d_2}$:

$$\overrightarrow{d_3} = \overrightarrow{d_1} + \overrightarrow{d_2} = \left((3t)i + 4(t^2 - 1)j\right) + (i - (e^t)j) = (3t + 1)i + (4(t^2 - 1) - e^t)j.$$

The velocity of the boat relative to ground

$$\vec{v} = \frac{d}{dt} \overrightarrow{d_3} = i \frac{d}{dt} (3t + 1) + j \frac{d}{dt} (4(t^2 - 1) - e^t).$$

Let's find components of the velocity due east and north:

$$v_{east} = \frac{d}{dt} (3t + 1) = 3 \frac{\mathrm{km}}{\mathrm{h}};$$ $$v_{north} = \frac{d}{dt} (4(t^2 - 1) - e^t) = (8t - e^t) \frac{\mathrm{km}}{\mathrm{h}}.$$

Answer: $3 \frac{\mathrm{km}}{\mathrm{h}}$ east and $(8t - e^t) \frac{\mathrm{km}}{\mathrm{h}}$ north.