A particle moves along the x axis according to the equation x = 2.08 + 3.02t - 1.00t, where x is in meters and t is in seconds.(a) Find the position of the particle at t= 2.80 s.18376m(b) Find its velocity at t- 2.80 s.-2.58v m/s(c) Find its acceleration at t = 2.80 s.-2v m/s?Need Help?Viewing Saved Work Revert to Last Response

Given Equation , x = 2.08 + 3.02t - 1t2 To find = a) position at t = 2.8 s b)…

A particle moves along the x axis according to the equation x 2.08 + 3.02t - 1.00t, where x is in meters and t is in seconds. (a) Find the position of the particle at t 2.80 s. 18376 (b) Find its velocity at t- 2.80 s. v m/s -2.58 (c) Find its acceleration at t 2.80 s. -2 v mys? Need Help? Read Master

A particle moves along the x axis according to the equation x 2.08 + 3.02t - 1.00t, where x is in meters and t is in seconds. (a) Find the position of the particle at t 2.80 s. 18376 (b) Find its velocity at t- 2.80 s. v m/s -2.58 (c) Find its acceleration at t 2.80 s. -2 v mys? Need Help? Read Master

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

Similar questions

A runner travels at the following speeds: A. 11.5 km/hr for 3.71 min. B. 10.6 km/hr for 2.39 min. C. 14.5 km/hr for 1.92 min. Determine the average velocity over the entire interval and report your result in km/hr.
An object has an initial velocity v = 2.4i + 4.5j m/s. Over a period of time t = 2.9 s, its velocity becames v2 = 2.4i +35 m/s. The magnitude of the object's acceleration (in m/s?) is: a. 1.73 b. 0.52 c. 2.59 d. 3.07 e. 0.26
At time t = 0, a particle has a velocity of m v = 4.00. The following graph shows the particles a(t) S acceleration vs. time. a. What was the particle's velocity at t = 3.00s ? 2 4 t b. What was the particle's instantaneous acceleration time(s) at t = 3.00s ? С. What was the average acceleration between t = 1.00s and t = 3.00s ? 4- 2. acceleration(m/s)
15. A ball is hit upward with an initial velocity of 128 feet per second. After t seconds, its height h(t) is given by the function h(t) = -16t?+128t+4. What is the maximum height in feet of the ball? a. 268 feet b. 256 feet C. 264 feet d. 260 feet
The Hint The position of a particle moving along the x axis is given by x = (21 +22t+8t²) m, where t is in s. What is the instantaneous velocity of the particle at t= 5.0 s ? a. b. C. e. f. 10.2m/s -2.0 m/s -82.0m/s 331 m/s 102m/s -6.0 m/s
A particle moves along the x axis according to the equation x = 1.91 + 3.01t – 1.00t2, where x is in meters and t is in seconds. (a) Find the position of the particle at t = 3.10 s. m (b) Find its velocity at t = 3.10 s. m/s (c) Find its acceleration at t = 3.10 s. m/s²
5. An object has velocity vo (t) = 3i + 1j m/s at time t = 0.00 s. The acceleration of the object is alt) = 2tî+ 1t2 f m/s². If the object is located at ro (t) = 11-1jm at time %3! t = 0.00 s, what is the position of the object at time t = 1.00 s?
2. A child running around in a field has the following position and velocity at time t = 0: x = 1.5 m, y = 2.0 m Vx = -3.5 m/s, vy=-1.0 m/s At time t = 2.5 s, it has the following position and velocity: x = -2.0 m, y = 1.5 m vx = -0.50 m/s, v, = 2.0 m/s Using this information, find the child's average velocity and average acceleration during that time interval. Write your answers in unit vector notation.
We are standing on the top of a 1040 feet tall building and launch a small object upward. The object's height, measured in feet, after t seconds is h(t) = 16t? + 128t + 1040. A) What is the object initial velocity? ft/second B) What is the highest point that the object reaches? feet
6. The position of a particle along a straight line is given by s = (1.5r - 13.52 + 22.5t) ft, where t is in seconds. Determine the position of the particle when t = 6 s and the total distance it travels during the 6- s time interval. Hint: Plot the path to determine the total distance traveled. Ans. s=6s = -27.0 ft ; Stot = 10.5 + 48.0+ 10.5 69.0 ft 7. A particle moves along a straight line such that its position is defined by s = ( - 6t + 5) m. Determine
A bus travels with a speed of 95 mi/h. This speed in S.I. units is (Given 1 mile = %3D .. 1.609 km) a. 212.55 m/s b. 42.46 m/s c. 342.00 m/s d. 26.39 m/s
A particle moves along the x axis according to the equation x = 2.10+ 2.96t - 1.00t2, where x is in meters and t is in seconds. (a) Find the position of the particle at t = 2.90 s. m (b) Find its velocity at t = 2.90 s. m/s (c) Find its acceleration at t = 2.90 s. m/s²