Explanation Given info. y = x − 1 Point location = ( 2 , 1 2 ) Concept used: V ( t ) = ( V ( x ) ) 2 + V y 2 Calculation: The parametric equation of the curve is x = t y = 1 t Therefore, the position vector of the particle can be described as r ( t ) = 〈 t , 1 t 〉 Velocity is the rate of change of position vector. V ( t ) = r 1 ( t ) = 〈 t , − 1 t 2 〉 Point of particle, ( 2 , 1 2 ) r ( t ) = 〈 t , 1 t 〉 Comparing both, you get t = 1 V ( t ) = 〈 1 , − 1 4 〉 The vector form of V ( t ) = i ∧ − 1 4 J ∧ Now unit vector V ∧ ( t ) = V →...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

See similar textbooks

Videos

Question

Chapter 14.3, Problem 20E

To determine

What is the velocity vector of a particle travelling the right along hyperbola $y = x^{- 1}$ with constant speed $5$ cm/s when the particle’s location is $(2, \frac{1}{2})$ ?

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.3, Problem 19E

Chapter 14.3, Problem 21E

Students have asked these similar questions

A particle moves along the top of the parabola y2 = 2x from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point (2, 2).

Assume a particle moves along the top branch of the parabola y2 = 3x from left to right going at a contant speed of 2 units/second. Determine the velocity of the particle as it moves through the point: (36/6, 6).

A particle moves along a straight line with equation of motion s=t^2 - 3t +2. Find the value of t at which the particle reverses its direction.

Chapter 14 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

What is the velocity vector of a particle travelling the right along hyperbola y = x − 1 with constant speed 5 cm/s when the particle’s location is ( 2 , 1 2 ) ?

Solution Summary: The author explains the velocity vector of a particle travelling the right along hyperbola y = x 1 with constant speed 5 cm/s.

Videos

What is the velocity vector of a particle travelling the right along hyperbola $y = x^{- 1}$ with constant speed $5$ cm/s when the particle’s location is $(2, \frac{1}{2})$ ?

Want to see the full answer?

Chapter 14 Solutions

What is the velocity vector of a particle travelling the right along hyperbola y = x − 1 with constant speed 5 cm/s when the particle’s location is ( 2 , 1 2 ) ?

Solution Summary: The author explains the velocity vector of a particle travelling the right along hyperbola y = x 1 with constant speed 5 cm/s.

Videos

What is the velocity vector of a particle travelling the right along hyperbola y=x−1 with constant speed 5 cm/s when the particle’s location is (2,12)?

Want to see the full answer?

Chapter 14 Solutions

What is the velocity vector of a particle travelling the right along hyperbola $y = x^{- 1}$ with constant speed $5$ cm/s when the particle’s location is $(2, \frac{1}{2})$ ?