Two particles leave the same point at the same time and both are moving along a horizontal line according to the law s= 4t3-2t2+3 and s= -7+28t-10t2, respctively. When are they moving at the same speed? Find their propositions, velocity and acceleration at that time?

Two particles leave the same point at the same time and both are moving along a horizontal line according to the law s= 4t3-2t2+3 and s= -7+28t-10t2, respctively. When are they moving at the same speed? Find their propositions, velocity and acceleration at that time?

Name: Two particles leave the same point at the same time and both are movin
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: Two particles leave the same point at the same time and both are moving along a horizontal line according to the law s= 4t3-2t2+3 and s= -7+28t-10t2, respctively. When are they moving at the same speed? Find their propositions, velocity and accelerat

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter6: Motion In Two Dimensions

Section: Chapter Questions

Problem 84A

See similar textbooks

Similar questions

If a particles velocity is given by the expression v(t) = 3.4t3 - 5.4t, What is the position of the particle after 5 seconds if it starts at rest?
A ball is thrown into the air by a baby alien on a planet in thesystem of Alpha Centauri with a velocity of 49 ft/s. Its heightin feet after t seconds is given by y = 49t −25t2.a.) Find the average velocity for the time period beginningwhen t0 = 1 second and lasting for the given time.t = .01 sec:t = .005 sec:t = .002 sec :t = .001 sec: b.) Estimate the instanteneous velocity when t = 1.
how much time would it take you to cover the61-m length of the walkway if, once you get on the walkway, youimmediately turn around and start walking in the opposite direction with a speed of 1.3 m>s relative to the walkway?
A balloon is ascending at the rate of 14m/s14m/s at a height of 98m98m above the ground where a packet is dropped from the balloon. After how much time and with velocity does it reach the ground?
Two objects, A and B, are moving along two different straight lines at constant speeds. With reference to a particular coordinate system in which distance is measured in metres, the position of A at time t (in minutes) is (3t−2,−2t+4), and the position of B is (−t−3,t−4). Let d be the distance between A and B at time t. Show that anexpression for d^2 in terms of t is given byd^2 =25t^2 −40t+65.
Two particles move along an x axis.The position of particle 1 is given by x = 6.00t2 + 3.00t + 2.00 (in meters and seconds); the acceleration of particle 2 is given by -8.00t (in meters per second squared and seconds) and, at t = 0, its velocity is 20 m/s.When the velocities of the particles match, what is their velocity?
If a ball is thrown upwards with velocity V1, maximum height attained by the ball is H1, if ball is thrown with V2 = 3V1, what will be the new maximum height?
The following table gives the position s1t2 of an object moving along a line at time t. Determine the average velocities over the time intervals 31, 1.014, 31, 1.0014, and 31, 1.00014. Then make a conjecture about the value of the instantaneous velocity at t = 1.
In a classic clip on Americas Funniest Home Videos, a sleeping cat rolls gently off the top of a warm TV set. Ignoring air resistance, calculate the position and velocity of the cat after (a) 0.100 s, (b) 0.200 s, and (c) 0.300 s.
The velocity of a partcile is given by the equation v = 5.0 t - 0.60 t2 (in m/s). At t=0, the particle is at rest. At what time will the particle be at rest again?
In an historical movie, two knights on horseback start from rest 96.9m apart and ride directly toward each other to do battle. Sir George’s acceleration has a magnitude of 0.213m/s^2, while Sir Alfred’s has a magnitude of 0.267 m/s^2. Relative to Sir Georges starting point, where do the knights collide?
A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 146 feet per second from an initial height of 89 feet off the ground, then the height of the projectile, hh, in feet, tt seconds after it's shot is given by the equation: h=−16t2+146t+89h=-16t2+146t+89 Find the two points in time when the object is 126 feet above the ground (to one decimal places). Answer: The object is 126 feet off the ground at the following times: (Enter your two answers separated by a comma. Round to ONE decimal places.)