Question 3: Vectors (13 points)
a. Draw the two vectors below added together. Then draw the sum of the two
vectors. (1 point)

Answer:

the coefficient of Kinetic friction between the tires and road is 0.38

Option A) .38 is the correct answer

Explanation:

Given that;

final velocity v = 0

initial velocity u = 15m/s

time taken t = 4 s

acceleration  a = ?

from the equation of motion        

v   =   u   +   at

we substitute

0 = 15 + a × 4

acceleration a = -15/4 =  - 3.75 m/s²    

the negative sign tells us that its a  deacceleration so the sign can be ignored.

Deacceleration due to friction a = μ × g

we substitute

3.75 = μ × 9.8    

μ = 3.75 / 9.8 = 0.3826 ≈ 0.38

Therefore the coefficient of Kinetic friction between the tires and road is 0.38

Option A) .38 is the correct answer

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

1) The force acting on the object during the time interval 0-3 seconds is 1/3 N.

2) The force acting on the object during the time interval 6-10 seconds is -0.5 N.

3) There is no time interval in which no force acts on the object.

(i) Force acting on the object in time interval 0-3 seconds. Force acting on the object is equal to the product of its mass and acceleration, i.e.,F = ma.

In the given velocity-time graph, the acceleration of the object can be determined by determining the slope of the velocity-time graph from 0 to 3 seconds.

Slope = (change in velocity) / (change in time)= (20-0) / (3-0) = 20/3 m/s^2

Acceleration, a = slope= 20/3 m/s^2

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × 20/3= 1/3 N.

Therefore, the force acting on the object during the time interval 0-3 seconds is 1/3 N.

(ii) Force acting on the object in time interval 6-10 seconds. Similar to the first question, the force acting on the object in time interval 6-10 seconds can be determined by determining the acceleration of the object during this time interval.

The slope of the velocity-time graph from 6 seconds to 10 seconds can be determined as follows:

Slope = (change in velocity) / (change in time)= (-20-20) / (10-6) = -40/4= -10 m/s^2 (negative sign indicates that the object is decelerating)

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × (-10)= -0.5 N.

Therefore, the force acting on the object during the time interval 6-10 seconds is -0.5 N.

(iii) Time interval in which no force acts on the object. There is no time interval in which no force acts on the object. This is because, as per Newton's first law of motion, an object will continue to remain in a state of rest or uniform motion along a straight line unless acted upon by an external unbalanced force.In other words, if the object is moving with a constant velocity, there must be a force acting on the object to maintain its motion.

Therefore, there is no time interval in which no force acts on the object.

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Given:

The resistance of copper is R = 9.28 ohms when the temperature is T = 0 degrees Celsius.

The temperature coefficient of resistivity is

To find the resistance when the temperature is T' = 24.97 degrees Celsius.

Explanation:

The resistance can be calculated as

Thus, the resistance is 10.28 ohms

An annotated bibliography is a list of citations to books, articles, and documents. Each citation is followed by a brief descriptive and evaluative paragraph, the annotation. The purpose of the annotation is to inform the reader of the relevance, accuracy, and quality of the sources cited.

The cue column is typically located on the left-hand side of the page and is used to jot down keywords or questions that serve as cues for recalling the main points of the lecture or reading. The note-taking area is located on the right-hand side of the page and is used to write down detailed notes about the lecture or reading.

The summary section is located at the bottom of the page and is used to summarize the key points of the notes. Overall, the Cornell method is an effective way to organize and retain information during lectures and readings.

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Answer:

the 2nd one is the right one

Explanation:

wave length is on top

The figure 2 correctly shows the part of the wave that represents the wavelength.

A wave is a dynamic disturbance that propagates and causes a change in equilibrium of one or more parameters in physics, mathematics, and related subjects. Quantities may oscillate regularly around an equilibrium (resting) value at certain frequency if a wave is periodic.

A travelling wave is one in which the entire waveform moves in one direction; in contrast, a standing wave is one in which two periodic waves are overlaid and move in the opposing directions.

In a standing wave, there are some points where the wave amplitude seems reduced or even zero, and these positions have null vibration amplitudes. A wave equation (standing wave field comprising two opposing waves) or a one-way wave equation (for single wave propagation in a certain direction) is frequently used to describe waves.

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We are given the following information

Mass = m = 45.0 kg

Initial speed = vi = 40.0 m/s

Time = t = 0.500 s

We are asked to find the average force that the seat belt exerted on her.

Recall from Newton's second law of motion,

Where a is the acceleration of the car and is given by

Where vf is the final velocity of the car that must be 0 since the car was stopped.

The negative sign indicates deacceleration since the card was stopped.

So, the force is

Therefore, the seat belt exerted a force of 3600 N on her.

Answer:

1) lack of time, 2) social influence, 3) lack of energy, 4) lack of willpower, 5) fear of injury, 6) lack of skill and 7) lack of resources

Explanation:

here are 7 barriers to physical activity choose 5

Answer:

1. Lack of time, injury, lack of energy, lack of willpower, and lack of confidence.

dssfbl;wihjkefdbvnfljedk.s;n.lasbcnk.esz

Answer:

3.6 kilometres per hour

Explanation:

for an approximate result, divide the speed value by 16.667

Explanation:

Using the equation of motion S = ut+1/2gt² where;

S is the height of the table = 0.60m

u is the initial velocity = 0m/s

g is the acceleration due to gravity = 9.81m/s²

t is the time taken

Substitute

0.6 = 0+1/2(9.81)t²

0.6 = 4.905t²

t² = 0.6/4.905

t² = 0.1223

t = √0.1223

t = 0.3497

t≈0.35secs

Hence it will take 0.35seconds for a ball launched off the edge of these tables to reach  the floor

Answer:

( About ) 6.8nC

Explanation:

We are given the equation |Q2| = mgd^2 / kQ1. Let us substitute known values into this equation, but first list the given,

Charge Q2 = +45nC = (45 × 10⁻⁹) C

mass of charge Q2 = 4.5 μg, force of gravity = 4.5 μg × 9.8 m/s² = ( 4.41 × 10^-5 ) N,

Distance between charges = 25 cm = 0.25 m,

k = Coulomb's constant = 9 × 10^9

_______________________________________________________

And of course, we have to solve for the magnitude of Q2, represented by the charge magnitude of the charge on Q2 -

(4.41 × 10^-5) = [(9.0 × 10⁹) × (45 × 10⁻⁹) × Q₂] / 0.25²

_______________________________________________________

Solution = ( About ) 6.8nC

The mass of the object is approximately 0.457 kg.

The mass of the object attached to the trolley can be calculated using the principle of conservation of momentum. Since the two trolleys couple together and move as a single system after the collision, the total momentum before and after the collision should be the same. Given the mass of one trolley is 0.80 kg and the initial speed is 1.1 m/s, the momentum before the collision is 0.80 kg * 1.1 m/s = 0.88 kg·m/s. After the collision, the total mass is the sum of the two trolleys, and the final speed is 0.70 m/s.

Using the momentum equation, the mass of the object can be calculated as follows:

Total momentum before collision = Total momentum after collision

0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s

Solving for the mass of the object, we get:

0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s

0.88 kg·m/s = 0.56 kg + 0.70 kg * mass of the object

0.88 kg·m/s - 0.56 kg = 0.70 kg * mass of the object

0.32 kg = 0.70 kg * mass of the object

Dividing both sides by 0.70 kg, we find:

mass of the object = 0.32 kg / 0.70 kg = 0.457 kg

The two trolleys collide and couple together, the total momentum before the collision is equal to the total momentum after the collision according to the principle of conservation of momentum.

The momentum of an object is defined as the product of its mass and velocity. In this case, the mass of one trolley is known (0.80 kg) and the initial speed is given (1.1 m/s), allowing us to calculate the momentum before the collision.

After the collision, the two trolleys move together at a new speed (0.70 m/s). By setting the initial momentum equal to the final momentum and solving for the unknown mass of the object, we can find its value.

In the calculation, we subtract the masses of the two trolleys from the total mass in order to isolate the mass of the object.

Dividing the difference in momentum by the product of the known mass and the new speed, we obtain the mass of the object. In this case, the mass of the object is approximately 0.457 kg.

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Figures 1, 2, and 4 are correct since they follow the condition of ohm's law.

Ohm's law states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit. The formula for Ohm's law is V=IR.

Ohm's law is I(the strength of the current flowing in a conductor)= V(the potential difference applied to the ends) divided by R(resistance)

so the figures have to satisfy the condition V = IR to demonstrate the ohmic behavior.

This indicates that current is directly proportional to voltage.

That is when current increases voltage also increases and vice versa.

Option 3 is incorrect, as when the current increases voltage cannot be constant.

Also, the graph should be linear:

Therefore, options 1, 2, and 4 are correct.

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Complete question:

Answer:

D

Explanation:

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

A force can make a body at rest to move. The magnitude of force parallel to surface 1 is 5.04 N and the magnitude of the force parallel to surface 2 is 4.02 N.

The force can be defined as the quantity which is expressed as the product of mass (m) and acceleration (a). It is known as the push or pull on an object which produces acceleration in the body on which it acts.

The equation which is used to calculate the force is given as:

F = ma

a) F₁ = m₁ g sin40

= 0.800 kg × 9.81 m/s² × 0.64

= 5.04 N

b) F₂ = m₂ g sin55

= 0.500 kg × 9.81 m/s² × 0.82

= 4.02 N

Thus the magnitude of the force parallel to surface 1 is 5.04 N and the magnitude of the force parallel to surface 2 is 4.02 N.

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Answer:

Explanation:

The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object, as shown in this formula. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function. Given a position vector →v=⟨a,b⟩, the magnitude is found by |v|=√a2+b2. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application.

The pressure applied by the standing book on the table is 1,800 N/m².

option B is the correct answer.

The pressure applied by the standing book on the table is determined from the ratio of weight of the book and the area of the standing book.

Mathematically, the formula for the pressure of a material is given as;

P = F / A

where;

The weight of the book , F = mg

where;

F = 1.5 kg  x  9.8 m/s²

F = 14.7 N

The dimension of the book include;

The height of the book is not in contact with the surface of the table, so the area of the book in contact with the table becomes;

A = w x b

W = 0.21 m  x  0.04 m

W = 8.4 X 10⁻³ m²

P = F / A

P = ( 14.7 N ) / ( 8.4 X 10⁻³ m² )

P = 1,750 N/m² ≈ 1,800 N/m²

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The interval between the first and second echo is 7 seconds. This means that after the initial sound wave reaches the first cliff, it takes a total of 7 seconds for the sound to travel to the second cliff and then return to the student as the second echo.

To determine the interval between the first and second echo, we need to consider the time it takes for sound to travel from the student to the first cliff, and then from the first cliff to the second cliff, and finally back to the student.

Let's break down the distances and calculate the time for each part of the journey:

Distance from the student to the first cliff: 330 meters

Time taken: t₁ = distance / speed = 330 m / 330 m/s = 1 second

Distance from the first cliff to the second cliff: 990 meters

Time taken: t₂ = distance / speed = 990 m / 330 m/s = 3 seconds

Distance from the second cliff back to the student: 990 meters

Time taken: t₃ = distance / speed = 990 m / 330 m/s = 3 seconds

Now, we can calculate the total interval between the first and second echo by adding up the individual times:

Interval between first and second echo = t₁ + t₂ + t₃ = 1 s + 3 s + 3 s = 7 seconds

Therefore, the interval between the first and second echo is 7 seconds. This means that after the initial sound wave reaches the first cliff, it takes a total of 7 seconds for the sound to travel to the second cliff and then return to the student as the second echo.

It's important to note that this calculation assumes a straight path for the sound waves and neglects factors such as air temperature and wind that can affect the speed of sound. Additionally, it assumes perfect reflection of sound waves off the cliffs, which may not be the case in real-world scenarios.

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Note the complete questions is:

A student stands 330m from a tall cliff which is 990m from another tall cliff. If the speed of sound between the cliffs is 330m/s.What is the interval between the first and second echo?

The temperature gradient in the flow of direction is 294525 W.

A temperature gradient is the gradual variance in temperature with distance. The slope of the gradient is consistent within a material. A gradient is established anytime two materials at different temperatures are in physical contact with each other.

Q= T/( L/ KA)

Q= ( 1500 − 450) / 0.15 / 9.35v * 4.35)

   = 294525 W

Units of measure of temperature gradients are degrees per unit distance, such as °F per inch or °C per meter.

Many temperature gradients exist naturally, while others are created. The largest temperature gradient on Earth is the Earth itself. Q= T/Ka.

Therefore, The temperature gradient in the flow of direction is 294525 W.

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Answer:

The truck will undergo the largest change in momentum if it has a greater mass than the small car.

Explanation:

The change in momentum of an object can be calculated using the equation:

Δp = m * Δv

where Δp represents the change in momentum, m represents the mass of the object, and Δv represents the change in velocity.

Since we are comparing the change in momentum of the car and the truck, we need to consider the masses of both vehicles.

Let's assume the mass of the car is represented by m_car, and the mass of the truck is represented by m_truck.

Since both vehicles collide head-on, the change in velocity (Δv) will be the difference between their initial velocities, considering that they are moving in opposite directions:

Δv = v_truck - v_car

Now, let's compare the change in momentum for the car and the truck:

For the car:

Δp_car = m_car * Δv

For the truck:

Δp_truck = m_truck * Δv

Comparing the magnitudes of the change in momentum, we can neglect the negative sign:

|Δp_car| = |m_car * Δv|

|Δp_truck| = |m_truck * Δv|

Since both Δv and Δp are positive values, we can conclude that the vehicle with the greater mass will undergo the largest change in its momentum.

Therefore, if the mass of the truck (m_truck) is greater than the mass of the car (m_car), then the truck will undergo the largest change in its momentum. Conversely, if the mass of the car is greater, then the car will undergo the largest change in its momentum.

(1) The distance of the image formed by the lens is determined as 6 cm.

(2) The image formed is real.

The distance of the image formed by the lens is calculated by applying the following formula as follows;

1/f = 1/v + 1/u

where;

The distance of the image formed by the lens is calculated as;

1/v = 1/f - 1/u

1/v = 1/2 - 1/3

1/v = 1/6

v = 6 cm

Since the sign of the image of the image is positive, the image formed is real.

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Answer:true

Explanation:

Option A) 3.9 km is the correct answer. the magnitude of Rhea's displacement vector is approximately 3.92 km.

In order to find out the magnitude of Rhea's displacement vector, we have to add up all of the displacement vectors.

Then we can use the Pythagorean theorem to calculate the magnitude of the resultant vector.

Since Rhea is first driving north for 2.4 km and then west for 3.1 km, we can represent her displacement vectors as follows: Δx = 0 km and Δy = 2.4 km for the first vector, and Δx = -3.1 km and Δy = 0 km for the second vector.

We can then add these vectors together by adding their components: Δx = 0 km + (-3.1 km) = -3.1 km and Δy = 2.4 km + 0 km = 2.4 km.

This gives us a resultant vector of -3.1 km east and 2.4 km north.

Using the Pythagorean theorem, we can find the magnitude of this vector: \(\sqrt{(\(-3.1 km)^{2} + (2.4 km)^{2} ) } = \sqrt{(9.61 + 5.76) km^{2} } = \sqrt{15.37 km^{2} } \approx 3.92 km.\)

Therefore, the magnitude of Rhea's displacement vector is approximately 3.92 km.

Therefore, option A) 3.9 km is the correct answer.

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The main difference between potential and kinetic energy is that one is the energy of what can be and one is the energy of what is. In other words, potential energy is stationary, with stored energy to be released; kinetic energy is energy in motion, actively using energy for movement.

1. The time taken for the object to reach the ground is 2.47 s

2. The speed with which the object will reach the ground is 26.206 m/s

1. How do I determine the time taken?

We can obtain the time taken for the object to reach the ground as      follow:

h = ½gt²

30 = ½ × 9.8 × t²

30 = 4.9 × t²

Divide both side by 4.905

t² = 30 / 4.9

Take the square root of both side

t = √(30 / 4.9)

t = 2.47 s

Thus, the time taken to reach the ground is 2.47 s

2. How do i determine the speed?

The speed the object will use to reach the ground can be obtained as follow:

v = u + gt

v = 2 + (9.8 × 2.47)

v = 2 + 24.206

v = 26.206 m/s

Thus, the speed is 26.206 m/s

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Answer:

It's the Second one Endocrine Cells

Explanation:

reflection ... . .......

Answer:

I will tell in order of pictures ok

Explanation:

pic1 transmission

pic2 reflection

pic3 refraction

pic4 absorbtion

Given:

the speed of the ball is

angle at which the ball is projected

Required: the velocity in the vector form

Explanation:

look at the diagram below

velocity in the vector form can be written as

here, i and j are unit vectors.

plugging all the value in the above relation, and solve

final answer is