QuestionFind the resultant velocity when crossing a river with a current of $15 \mathrm{~m/s}$ downstream and crossing at $2 \mathrm{~m/s}$ north.

Studdy Solution

STEP 1

Assumptions1. The current of the river is15 m/s downstream. . You are crossing the river at m/s north.
3. The downstream direction is considered as the positive x-axis and the north direction is considered as the positive y-axis.
4. The resultant velocity is the vector sum of the velocity due to the current of the river and the velocity of crossing the river.

STEP 2

We first represent the velocities as vectors. The velocity due to the current of the river is represented as a vector along the x-axis and the velocity of crossing the river is represented as a vector along the y-axis.
$\vec{v}_{current} =15 \hat{i} \, m/s$ $\vec{v}_{crossing} =2 \hat{j} \, m/s$

STEP 3

The resultant velocity is the vector sum of the two velocities. We can find this by adding the two vectors.
$\vec{v}_{resultant} = \vec{v}_{current} + \vec{v}_{crossing}$

STEP 4

Substitute the values of the two velocities into the equation.
$\vec{v}_{resultant} =15 \hat{i} +2 \hat{j} \, m/s$

STEP 5

The magnitude of the resultant velocity can be found using the Pythagorean theorem.
$|\vec{v}_{resultant}| = \sqrt{(v_{x})^2 + (v_{y})^2}$

STEP 6

Substitute the x and y components of the resultant velocity into the equation.
$|\vec{v}_{resultant}| = \sqrt{(15)^2 + (2)^2} \, m/s$

STEP 7

Calculate the magnitude of the resultant velocity.
$|\vec{v}_{resultant}| = \sqrt{225 +4} \, m/s = \sqrt{229} \, m/s \approx15.13 \, m/s$ The resultant velocity is approximately $15.13 \, m/s$ in the direction of the vector $\vec{v}_{resultant} =15 \hat{i} +2 \hat{j}$ .

Was this helpful?

QuestionFind the resultant velocity when crossing a river with a current of $15 \mathrm{~m/s}$ downstream and crossing at $2 \mathrm{~m/s}$ north.

Studdy Solution

STEP 1

STEP 2

We first represent the velocities as vectors. The velocity due to the current of the river is represented as a vector along the x-axis and the velocity of crossing the river is represented as a vector along the y-axis.
$\vec{v}_{current} =15 \hat{i} \, m/s$ $\vec{v}_{crossing} =2 \hat{j} \, m/s$

STEP 3

The resultant velocity is the vector sum of the two velocities. We can find this by adding the two vectors.
$\vec{v}_{resultant} = \vec{v}_{current} + \vec{v}_{crossing}$

STEP 4

Substitute the values of the two velocities into the equation.
$\vec{v}_{resultant} =15 \hat{i} +2 \hat{j} \, m/s$

STEP 5

The magnitude of the resultant velocity can be found using the Pythagorean theorem.
$|\vec{v}_{resultant}| = \sqrt{(v_{x})^2 + (v_{y})^2}$

STEP 6

Substitute the x and y components of the resultant velocity into the equation.
$|\vec{v}_{resultant}| = \sqrt{(15)^2 + (2)^2} \, m/s$

STEP 7

Calculate the magnitude of the resultant velocity.
$|\vec{v}_{resultant}| = \sqrt{225 +4} \, m/s = \sqrt{229} \, m/s \approx15.13 \, m/s$ The resultant velocity is approximately $15.13 \, m/s$ in the direction of the vector $\vec{v}_{resultant} =15 \hat{i} +2 \hat{j}$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the resultant velocity when crossing a river with a current of 15 m/s15 \mathrm{~m/s}15 m/s downstream and crossing at 2 m/s2 \mathrm{~m/s}2 m/s north.

Studdy Solution

STEP 1

STEP 2

STEP 3

The resultant velocity is the vector sum of the two velocities. We can find this by adding the two vectors.v⃗resultant=v⃗current+v⃗crossing\vec{v}_{resultant} = \vec{v}_{current} + \vec{v}_{crossing}vresultant​=vcurrent​+vcrossing​

STEP 4

Substitute the values of the two velocities into the equation.v⃗resultant=15i^+2j^ m/s\vec{v}_{resultant} =15 \hat{i} +2 \hat{j} \, m/svresultant​=15i^+2j^​m/s

STEP 5

The magnitude of the resultant velocity can be found using the Pythagorean theorem.∣v⃗resultant∣=(vx)2+(vy)2|\vec{v}_{resultant}| = \sqrt{(v_{x})^2 + (v_{y})^2}∣vresultant​∣=(vx​)2+(vy​)2​

STEP 6

Substitute the x and y components of the resultant velocity into the equation.∣v⃗resultant∣=(15)2+(2)2 m/s|\vec{v}_{resultant}| = \sqrt{(15)^2 + (2)^2} \, m/s∣vresultant​∣=(15)2+(2)2​m/s

STEP 7

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the resultant velocity when crossing a river with a current of $15 \mathrm{~m/s}$ downstream and crossing at $2 \mathrm{~m/s}$ north.

The resultant velocity is the vector sum of the two velocities. We can find this by adding the two vectors.
$\vec{v}_{resultant} = \vec{v}_{current} + \vec{v}_{crossing}$

Substitute the values of the two velocities into the equation.
$\vec{v}_{resultant} =15 \hat{i} +2 \hat{j} \, m/s$

The magnitude of the resultant velocity can be found using the Pythagorean theorem.
$|\vec{v}_{resultant}| = \sqrt{(v_{x})^2 + (v_{y})^2}$

Substitute the x and y components of the resultant velocity into the equation.
$|\vec{v}_{resultant}| = \sqrt{(15)^2 + (2)^2} \, m/s$