### Problem DescriptionA particle moving in the x-y plane has a velocity at time \( t = 5.25 \, \text{s} \) given by \( (3.9\mathbf{i} + 6.0\mathbf{j}) \, \text{m/s} \), and at \( t = 5.34 \, \text{s} \) its velocity has become \( (4.05\mathbf{i} + 6.13\mathbf{j}) \, \text{m/s} \). Calculate the magnitude \( a \) of its average acceleration during the 0.09-s interval and the angle \( \theta \) it makes with the x-axis.### Answers:\[ a = \, \boxed{} \, \text{m/s}^2 \]\[ \theta = \, \boxed{} \, ^\circ \]This problem involves determining the average acceleration of a particle given its velocity at two different points in time. The necessary steps involve using the change in velocity over time to find the average acceleration vector and then calculating its magnitude and direction.

Name: ### Problem DescriptionA particle moving in the x-y plane has a veloc
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: ### Problem DescriptionA particle moving in the x-y plane has a velocity at time \( t = 5.25 \, \text{s} \) given by \( (3.9\mathbf{i} + 6.0\mathbf{j}) \, \text{m/s} \), and at \( t = 5.34 \, \text{s} \) its velocity has become \( (4.05\mathbf{i} +

Write the given values with the suitable variables. t1=5.25 su=3.9i+6j m/st2=5.34 sv=4.05i+6.13j m/s…

Science

Physics

A particle moving in the x-y plane has a velocity at time t = 5.25 s given by (3.9i+ 6.0j) m/s, and at t = 5.34 s its velocity has become (4.05i + 6.13j) m/s. Calculate the magnitude a of its average acceleration during the 0.09-s interval and the angle it makes with the x- axis. Answers: i m/s² a= e- i

A particle moving in the x-y plane has a velocity at time t = 5.25 s given by (3.9i+ 6.0j) m/s, and at t = 5.34 s its velocity has become (4.05i + 6.13j) m/s. Calculate the magnitude a of its average acceleration during the 0.09-s interval and the angle it makes with the x- axis. Answers: i m/s² a= e- i

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the…

A: Write the given values with suitable variables. at=-0.032 m/s315 s-tx0=-14 mv0x0=7 m/st=10 s

Q: A robot has been designed and is tested on the course above (A-B-C-D-E). If the robot takes 3.4…

A: Required to find the average speed of the robot.

Q: The eguation r(t)= (3t+ 9) i + (8t - 2) j+(4t) k is the position of a particle in space at time t.…

A: Introduction: Kinematics helps to find out the velocity, acceleration of the particle at any time t…

Q: A spaceship is traveling at a velocity of v0 = (37.3 m/s)i when its rockets fire, giving it an…

A: Data Given , Initial velocity ( V0 ) = ( 37.3 i ) m/s Acceleration ( a ) = ( 2.55 i + 4.19 k ) m/s2…

Q: The position of a particle is: r(t) = [(4.62 m/s^2) t^2] i + (-3.38m) j + [(4.04 m/s^3) t^3] k…

Q: The acceleration of an object (in m/s2) is given by the function a(t)=5sin(t)a(t)=5sin(t). The…

Q: A particle starts from the origin with velocity 91 m/s at t = 0 and moves in the xy plane with a…

Q: A rocket is fired from rest at x = 0 and travels along a parabolic trajectory described by…

Q: A spacecraft has an initial velocity V; =(-32.2 m/s) î+(20.3 m/s)ĵ. The thrusters are fired, and the…

Q: An object has position given by the vector r = [3.0m + (6.0m/s)t] i + [2.0m - (2.0m/s2)t2] j, where…

Q: A bottle rocket is launched straight up. Its height in feet (y) above the ground x seconds after…

A: Given: The equation of motion is given as, y=-16x2+124x+8

Q: A particle P starts at time t = 0 s at the position x = -13 m y = -10 m. The velocity of the…

Q: The distance s(t), in meters, is given as a function of time, in seconds, below.…

Q: n object's velocity as a function of time in one dimension is given by the expression; v(t) = 2.71t…

A: Object's velocity as a function of time is Find:Object's velocity at time

Q: A sports car moving at constant velocity travels 130m in 4.9s. It then brakes and comes to a stop in…

A: First we find the initial speed of the car then we apply equation of motion to find the acceleration…

Q: A particle leaves the origin with an initial velocity = (7.83î) m/s and a constant acceleration à =…

Q: The velocity of a particle moving in the x-y plane is given by (7.15i +4.12j) m/s at time t = 5.37…

A: We are given velocity of particle. We are also given average acceleration. We hence find the final…

Q: 1. A particle initially located at the origin has a constant acceleration of a = 3.00j m/s and an…

A: Given data The magnitude of the constant acceleration is a = 3 j m/s2 The initial velocity of the…

Q: At time t = 16 s, the velocity of a particle moving in the x-y plane is v = 0.12i + 2.69j m/s. By…

Q: vo-dimensional motion: An object has a position given by r = [2.0 m + (3.00 m/s)t]i + [3.0 m - (2.00…

Q: A small object moves along the xx-axis with acceleration ax(t) = − (0.0320m/s3)(15.0s−t) −…

A: The acceleration of small object moving along x axis is given by : a(t) = - (0.0320 ms3) (15s-t) the…

Q: (a) Find the velocity and position vectors as functions of time t. m/s = m (b) Find the equation of…

A: a) The particle’s velocity can be determined as, The particle’s velocity can be determined as,

Q: At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of v, = (3.00…

Q: An object undergoes acceleration 2.3i +3.6j m/s² for 5.9 s at which time its velocity is 33i +15j…

Concept explainers

Projectile Motion

Topic Video

Question

### Problem Description

A particle moving in the x-y plane has a velocity at time \( t = 5.25 \, \text{s} \) given by \( (3.9\mathbf{i} + 6.0\mathbf{j}) \, \text{m/s} \), and at \( t = 5.34 \, \text{s} \) its velocity has become \( (4.05\mathbf{i} + 6.13\mathbf{j}) \, \text{m/s} \). Calculate the magnitude \( a \) of its average acceleration during the 0.09-s interval and the angle \( \theta \) it makes with the x-axis.

### Answers:
\[
a = \, \boxed{} \, \text{m/s}^2
\]

\[
\theta = \, \boxed{} \, ^\circ
\]

This problem involves determining the average acceleration of a particle given its velocity at two different points in time. The necessary steps involve using the change in velocity over time to find the average acceleration vector and then calculating its magnitude and direction.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

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

Similar questions

(a) The velocity of a particle changes from v1 = 2i + 5j -3k m/s to v2 = 5i - 7j+ 2k m/s in 2 s. What is its average acceleration? (b) A particle has an acceleration of a = -7i + 3j m/s2 for a period of 4 s. After this time the velocity is V2 = 6i - 2k m/s. What was the initial velocity?
An object's position in the as a function of time obeys the equation; x (t) = 3t – 2t³ + 1.5t where all constants have proper SI Units. What is the speed of the object in the x direction at t = 1.75 seconds?
An object's position as a function of time in one dimension is given by the expression; 3.89t2 + 2.22t + 7.48 where are constants have proper SI Units. What is the object's average velocity between the times t = 3.16 s and t = 8.38 s?
A particle moves along the positive x-axis and its position as a function of time is x(t)=(19.9 m/s)t − (1.7 m/s^2)t^2. What is the average velocity in the time interval between 2.3s and 3.7s in m/s?
An object is travelling in 1-dimension (i.e. along the x- axis) with a velocity described by the equation: v(t)=v0+1/6st^3 , where s and v0 are constants. The object’s position, and acceleration at time t = 0 are given by d0, and a0 respectively. In terms of the variables given, what is the average acceleration of the object over the interval from 0 to 2 seconds.
A particle starts from the origin at r=0 with a velocity of 8.0 m/s and moves in the xy plane with constant acceleration (3.5i +1.7)) m/s1. (a) When the particle's x coordinate is 27 m, what is its y coordinate? (b) When the particle's x coordinate is 27 r. what is its speed? m/s
An object's position as a function of time in one dimension is given by the expression; 3.89t2 + 2.22t + 7.48 where are constants have proper SI Units. What is the object's average velocity between the times t = 3.16 s and t = 8.38 s? with everything in 3 sig figs
A particle has a velocity of v= (3i+5j)m/s and a constant acceleration of a=(-1.5i-4.5j)m/s^2 What is its velocity after 2 seconds? Is it moving faster or slower after 2 seconds?
An object moves on xy plane according to the equation 7 = 20 cos(3t) î + (1,5t + 3,5t2)j. Determine its position at t = 0. Find its average velocity and average acceleration between t = 2s and t = 5s.
An unidentified flying object (UFO) is observed to travel a total distance of 49000 m, starting and ending at rest, over a duration of 3.07 s. Assuming the UFO accelerated at a constant rate to the midpoint of its journey and then decelerated at a constant rate the rest of the way, what was the magnitude of its acceleration? Express your answer in g s , where 1 g = 9.81 m/s^2. 20,796 g s 10,398 g s 2,120 g s 1,060 g s
7:31 a 4 O 0令ll回 10. The displacement of the object between t = 0 and t = 4 s is: (A) 22 m (B) 28 m (C) 40 m (D) 42 m (E) 60 m Velocity (mis) Time (b a e 11. The above graph represents the velocity as a function of time of a moving object. What is the net displacement between t = 0 s and t = 4 s? (A) Om (B) 16 m (C) 24 m (D) 36 m (E) 48 m Acceleration vs Time Time (s) a
A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 4.10 m/s. What is the x-coordinate of the object when t = 10.0 s?