Velocity and acceleration of a particle at time t = 0 are u = (2î + 3j) m/s and a = (4î +2j) m/s? respectively. Find the velocity and displacement of particle at t = 2 s.
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- The coordinates of a particle moving along a curve are x(t) = -2t2 +15 and y(t) = t2 -10t + 15 , m whent is in seconds. Calculate the magnitude of velocity and acceleration when t= 5 sec.A particle moves along the x-axis according to the equation x 2.00 + 3.00t - the acceleration of the particle. 1.00t2. where x is in meters and t is in seconds. At t=D3.00s, findThe acceleration of a particle is given by a 4t - 37, where a is in meters per second squared and tis in seconds. Determine the velocity and displacement as functions of time. The initial displacement at t = O is so = 9 m, and the initial velocity is vo = 0 m/s. After you have the general expressions, evaluate these expressions at the indicated times. Answers: m, v m/s Att- 2.9s. m. v- m/s Att 10.6 s, S-
- A particle moves along the x-axis so that its coordinate follows the law x(t) =2t^2 +4, where x is in meters, and t is in seconds. Find the particles velocity (in m/s) O 23 O 7t O. 17.3 4t O none of the aboveU, m/s Find the displacement of the particle at t=4 sec starting at us = 40 is at the origion. 40 t,s 4.0A particle is thrown into air at the origin with initial velocity v = Ai + Bj where A and B are positive constants. The acceleration of the particle is given by -ati – yj where a and y are positive constants, and t is time. Find the range of motion assuming that it lands on ground level again. 4AB 2α Β3 37 3 2α Β3 3 3 2AB 3y 2AB 4a B³ 3 3 2AB a B³ - 3y 3 y3 4AB 2α Β3 - 37 3 73
- The acceleration of a particle is given by a = 2t-20, where a is in meters per second squared and t is in seconds. Determine the velocity and displacement as functions of time. The initial displacement at t = 0 is so = -5 m, and the initial velocity is vo = 6 m/s. Once you have determined the functions of time, answer the questions. Questions: When t = 5.1 s, S= V = a = Mi MI i m m/s m/s²Taking north to be the positive y direction and east to be the positive x direction, a particle's position is given byr(t) = (20.2m/s)t^i+(5.15 m/s2)t2^jAt what time, in seconds, is the particle traveling exactly northeast?Chapter 02, Problem 10 In reaching her destination, a backpacker walks with an average velocity of 1.26 m/s, due west. This average velocity results, because she hikes for 6.62 km with an average velocity of 2.54 m/s due west, turns around, and hikes with an average velocity of 0.381 m/s due east. How far east did she walk (in kilometers)? dw VE West East dɛ Number Units the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work
- The velocity of a particle which moves along the s-axis is given by v=8+10+/7 where t is in seconds and vis in meters per second. Evaluate the displacements, velocity v. and acceleration a when t = 5.6s. The particle is at the origins=0 when t = 0. Answers: Att = 5.6s, S= V= a= i m m/s m/s²A particle moves along the x axis according to the equation x=2.00 + 3.00t - 1.00t², where x is in meters and t is in seconds. At t=3.00s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration.A particle moves along two dimensions based on the following position vector r= [2.0 m + (3.00 m/s)f ]i + [(3.0 m)t - (2.00 m/s2)rlj a. Find the distance it covered in the first minute. b. Find the general expression of the instantaneous acceleration of the particle.