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Velocity vectors are the result of combining two terms: velocity and vectors. The rate of change of displacement is what we call velocity and Vectors are physical quantities that have both a magnitude and a direction. Vectors are graphically represented by a line with an arrowhead on it. In this method, we split the diagonal velocity V into horizontal vx and vertical vy components to make the calculations easier.
Also Read: Velocity unit
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Key Takeaways: Velocity Vectors, Components, Magnitude, Angle of total Velocity
Velocity Vectors
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Since velocities can point in diagonal directions, two-dimensional Motion is unquestionably more complex than one-dimensional motion. Consider the case of a baseball going horizontally and vertically around the same time with a diagonal velocity of v to better grasp this statement. We divide this velocity component into horizontal vx and vertical vy components to make our calculations easier.
Velocity Vector
We may deal with each direction separately by splitting the diagonal velocity v into horizontal vx and vertical vy components. Even when the vector is something other than velocity, such as force or momentum, this method of splitting up the vectors into components works.
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| Related Topics | ||
|---|---|---|
| Drift Velocity | Acceleration and Velocity | Velocity Selector |
| Derivation of Escape Velocity | Velocity Vectors | Derivation of Drift Velocity |
Vector Components
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Let’s go through the fundamental trigonometric rules before we get started with velocity vectors:
The following relationship can be used to connect the side length hypotenuse, opposing, adjacent, and one of the angles as shown below.
Trigonometry Rules
When we split the diagonal vector into two perpendicular components, the total vectors and their components form a right-angle triangle. As a result, the same trigonometric laws apply to velocity vectors. The following diagram illustrates the relationship between the trigonometric rule and velocity vectors:
Before we deal with velocity vectors, let us recall the following trigonometric rules:
Vector Components
We can see that Vx is regarded as the adjacent side, Vy as the opposite, and v as the hypotenuse.
Read More: Escape Velocity and Orbital Velocity
Applications of Velocity Vectors
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Here written are some applications of velocity vectors:
- The rate of change of an object's position is represented by a velocity vector.
- The magnitude of a velocity vector indicates an object's speed, whereas the vector direction indicates the object's direction.
- The principles of vector addition can be used to add or subtract velocity vectors.
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Things to Remember
- The rate of change of an object’s position is represented by a velocity vector.
- The magnitude of a velocity vector represents the speed of an action, whereas the vector direction signifies the direction of an object.
- To calculate velocity, divide the distance by the time it takes to traverse the same distance, then multiply by the direction.
- Force, velocity, acceleration, displacement, and momentum are examples of vector quantities.
- A negative velocity object is the one that works in the opposite direction. When an object’s speed slows down, its acceleration vector works in the different direction of its motion.
Read More: Energy Orbiting Satellites
Sample Questions
Ques: A ball is thrown at a 35° angle to the horizontal with an initial velocity of 70 feet per second. Calculate the velocity’s vertical and horizontal components. (3 Marks)
Ans: Let’s take v for velocity, and write v in unit vector form using the following information:
V =70 (cos(35°))i +70 (sin(35°))j
Simplify the scalars, we get:
V ≈ 57.34i + 40.15j
As the scalars are the horizontal and vertical components of v ,
The horizontal component is therefore 57.34 feet per second, whereas the vertical component is 40.15 feet per second.
Ques: A man is walking due east at a rate of 5km/h and the rain appears to be falling vertically with a speed of 12km/h. Calculate the direction of rain. (3 Marks)
Ans: VRain, Person =VRain−VPerson =−12j^ (vertically down)
Thus, VRain−(5i^)=−12j^
or,
VRain=5i^−12j^
Thus, the direction is tanθ¹(12/5)=22degree37′ east of vertical (+5,−12) means 4th quadrant, thus east of vertical.
Ques: A 4 mph eastbound motorboat faces a 7.0 mph northbound current. (5 Marks)
1) What is the motorboat’s resultant velocity?
2) If the river is 80 metres wide, how long does it take for the boat to travel from shore to shore?
3) How far downstream does the boat travel before reaching the opposite shore?
Ans: The resultant velocity can be calculated with the use of Pythagorean theorem. The resultant is the hypotenuse of a right triangle with sides of 4 m/s and 7 m/s. It is
- √[ (4 m/s)² + (7 m/s)² ] = 8.06 m/s
Its direction can be found by using a trigonometric function.
Direction = tanθ¹ [ (7 m/s) / (4 m/s) ] = 60°
- The time to cross the river is t = d / v = (80 m) / (4 m/s) = 20 s
- The distance travelled downstream is d = v X t = (7 m/s) X (20 s) = 140 m
Ques: A motorboat moving east at 5 mph confronts a current travelling south at 2.5 mph. (5 Marks)
1) What is the motor boat’s resultant velocity?
2) If the river is 80 metres wide, how long does it take for the boat to travel from shore to shore?
3) How far downstream does the boat travel before reaching the opposite shore?
Ans:
- The resultant velocity can be determined by using the Pythagorean theorem. The resultant is the hypotenuse of a right triangle with sides of 5 m/s and 2.5 m/s. It is
√ [ (5 m/s)² + (2.5 m/s)² ] = 5.59 m/s
Its direction can be found by using a trigonometric function.
Direction = 360 degrees - invtan[ (2.5 m/s) / (5 m/s) ] = 333.4 degrees
NOTE: the direction of the resultant velocity (like any vector) is represented as the counterclockwise angle of rotation from due East.
- The time to cross the river is t = d / v = (80 m) / (5 m/s) = 16.0 s
- The distance travelled downstream is d = v • t = (2.5 m/s) • (16.0 s) = 40 m
Ques: A speedboat moving east at 6 mph confronts a current travelling south at 3.8 mph. (5 Marks)
1) What is the motor boat’s resultant velocity?
2) If the river is 120 metres wide, how long does it take for the boat to travel from shore to shore?
Ans: The Pythagorean theorem can be used to calculate the resulting velocity. The hypotenuse of a right triangle with sides of 6 m/s and 3.8 m/s is the resultant. It is,
- √ [ (6 m/s)² + (3.8 m/s)² ] = 7.10 m/s
Its direction can be determined using a trigonometric function.
Direction = 360 degrees - invtan[ (3.8 m/s) / (6 m/s) ] = 327.6 degrees
NOTE: the direction of the resultant velocity (like any vector) is expressed as the counterclockwise direction of rotation from due East.
- The time to cross the river is t = d / v = (120 m) / (6 m/s) = 20.0 s
Ques: A plane can travel with a speed of 80 mi/hr with respect to the air. Determine the resultant velocity of the plane (magnitude only) if it encounters a (5 Marks)
a) 10 mi/hr headwind.
b) 10 mi/hr tailwind.
c) 10 mi/hr crosswind.
d) 60 mi/hr crosswind.
Ans: a) A headwind would decrease the resultant velocity of the plane to 70 mi/hr.
b) A tailwind would increase the resultant velocity of the plane to 90 mi/hr.
c) 10 mi/hr crosswind would increase the resultant velocity of the plane to 6 mi/hr.
This can be determined using the Pythagorean theorem:
√[ (80 mi/hr)² + (10 mi/hr)² ])
d) A 60 mi/hr crosswind would increase the resultant velocity of the plane to 100 mi/hr.
This can be determined using the Pythagorean theorem:
√[ (80 mi/hr)² + (60 mi/hr)² ] )
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