(a)
The components of the particle’s velocity at
(a)
Answer to Problem 5P
The
Explanation of Solution
Formula to calculate net force acts on a particle is,
Here,
Formula to calculate acceleration of a particle is,
Here,
Substitute
Substitute
Formula to calculate the velocity of the particle is,
Here,
Substitute
Conclusion:
Therefore, the
(b)
The direction of the motion of the particle at
(b)
Answer to Problem 5P
The direction of the particle’s motion at
Explanation of Solution
Formula to calculate the direction of the moving particle is,
Here,
Substitute
Conclusion:
Therefore, the direction of particle’s motion at
(c)
The displacement of the particle during
(c)
Answer to Problem 5P
The displacement of the particle during
Explanation of Solution
Formula to calculate the displacement of the particle is,
Here,
Substitute
Conclusion:
Therefore, the displacement of the particle during
(d)
The coordinates of the particle at
(d)
Answer to Problem 5P
The coordinates of the particle at
Explanation of Solution
The initial position of the particle is
The final coordinates of the particle at
Conclusion:
Therefore, the coordinates of the particle at
Want to see more full solutions like this?
Chapter 4 Solutions
Principles of Physics: A Calculus-Based Text
- According to a simplified model of a mammalian heart, at each pulse approximately 20 gg of blood is accelerated from 0.22 m/sm/s to 0.36 m/sm/s during a period of 0.11 ss . What is the magnitude of the force exerted by the heart muscle?arrow_forwardAccording to a simplified model of a mammalian heart, at each pulse approximately 20 g of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?arrow_forwardA particle of mass 1.15 kg is subject to a force that is always pointed towards the East but whose magnitude changes linearly with time t. The magnitude of the force is given as F = (6 N/s)t. Let the x-axis point towards the East. (a) Determine the change in the velocity Δv, in meters per second, of the particle between t = 0 and t = 1.2 sec. (b) Determine the change in x-coordinate Δx, in meters, of the particle between t = 0 and t = 1.2 if the initial velocity is 12.6 m/s, and points in the same direction as the force.arrow_forward
- A 2.0 kg projectile with initial velocity v→ = 6.5i m/s experiences the variable force F→ = -2.0ti +4.0t2j N, where t is in s. What is the projectile's speed at t = 2.0 s? Also, at what instant of time is the projectile moving parallel to the y-axis?arrow_forwardTwo forces, F1 = (6.10î + 4.30ĵ) N and F2 = (3.40î + 7.65ĵ) N, act on a particle of mass 2.20 kg that is initially at rest at coordinates (+2.05 m, −4.50 m). (a) What are the components of the particle's velocity at t = 10.7 s? v= (in m/s)arrow_forwardA runner of mass 62.0 kg initially moves at a speed of 8.05 m/s.How long must an average external force of 1.60* 10^2 N act to bring the runner to rest?arrow_forward
- A wood block weighing 44.75 N rests on a rough horizontal plane, the coefficient of friction being 0.40.If a bullet weighing 0.25 N is fired horizontally into the block with a velocity of 600 m/s, how far will theblock be displaced from its initial position ? Assume that the bullet remains inside the block. A.1.41m B.0.89m C.2.06m D.1.98marrow_forwardTwo forces, F1 = (3.90î − 2.60ĵ) N and F2 = (3.65î − 8.05ĵ) N, act on a particle of mass 2.10 kg that is initially at rest at coordinates (+1.60 m, +3.50 m). (a) What are the components of the particle's velocity at t = 11.8 s? v = 42.362i−59.826j m/s (b) In what direction is the particle moving at t = 11.8 s? 305.30 2 (c) What displacement does the particle undergo during the first 11.8 s?Δr = 863.07 m is not correct answer (d) What are the coordinates of the particle at t = 11.8 s? x = 249 m y = -713.5 m is not correct answer Please give correct, complete, and detailed formulas for correctly solving these questions.arrow_forwardA 3.5 kg particle moves along an x axis according to x(t) = −12 + 5t2 − 4t3, with xin meters and t in seconds. In unit-vector notation, what is the net force acting on theparticle at t = 6 s?arrow_forward
- A loaded tractor-trailer with a total mass of 5500 kg traveling at 2.5 km/h hits a loading dock and comes to a stop in 0.56 s . What is the magnitude of the average force exerted on the truck by the dock?arrow_forwardA particle of mass 1.06 kg begins at rest and is then subject to a force in the positive x direction that changes with time as given by the following function: F = mg[1-e-2.1t ], where g is the acceleration due to gravity. Part (a) Determine the change in the velocity Δv of the particle between t = 0 and t = 2.4 sec. Part (b) Determine the change in x-coordinate of the particle Δx between t = 0 and t = 2.4.arrow_forwardA particle of mass 2.4 kg is subject to a force that is always pointed towards the North but whose magnitude changes quadratically with time. Let the y-axis point towards the North. The magnitude of the force is given as F = 6t2, and has units of newtons Determine the change in velocity Δv, in meters per second, of the particle between t=0 and t=1.3s. Determine the change in y coordinate, in meters, of the particle Δ y betweent=o and t=1.3 if the initial velocity= 15.9 m/s and directed North, in the same direction as the force.arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning