Answered: A particle leaves the origin with an… | bartleby

Related questions

Question

A particle leaves the origin with an initial velocity ý = (7.46i) m/s and a constant acceleration á = (-3.97i – 1.95j) m/s². when the particle reaches its maximum x coordinate, what are (a) its velocity, (b)
its position vector?
(a) Number
Units
(b) Number
i+
lj Units

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

Please asap
Suppose the position vector for a particle is given as a function of time by , with x(t) = at + b and y(t) =ct2 + d, where a = 1.00 m/s, b = 1.00 m, c= 0.125 m/s2, and d = 1.00 m. (a) Calculate the average velocity during the time interval from t = 1.00 s to t = 4.00 s. (b) Determine the velocity and the speed at t = 3.00 s. (c) Determine the acceleration at t = 3.00 s.
I need help here please
The position vector 4.70t|i+ [et + ft2 locates a particle as a function of time t. Vector is in meters, t is in seconds, and factors e and f are constants. The figure gives the angle e of the particle's direction of travel as a function of time (0 is measured from the positive direction of the x axis). What are (a) factor e and (b) factor f, including units? 20° 0° 10 20 -20° t (s) (a) Number Units (b) Number Units
A particle leaves the origin with an initial velocity 3.00î m/s and a constant acceleration ở -1.00î – 0.50ĵ m/s². When it reaches its maximum x coordinate, what are its (a) velocity and (b) position vector?
A proton initially has 7 = (-7.10)î + (5.10)Ƒ + (−6.60)î : and then 2.70 s later has 7 = (4.70)î + (5.10)ĵ + (13.0)ê (in avg meters per second). (a) For that 2.70 s, what is the proton's average acceleration a in unit vector notation, (b) in magnitude, and the angle between a and the positive direction of the x axis? avg (a) Number i 11.8 (b) Number i (c) Number i î+ Units Units i 19.6 k Units m/s
A snowmobile is originally at the point with position vector 30.5 m at 95.0° counterclockwise from the x-axis, moving with velocity 4.89 m/s at 40.0°. It moves with a constant acceleration 1.72 m/s2 at 200°. After 5.00 s have elapsed, find the following. (Express your answers in vector form.) (a) its velocity vector. (b) its position vector.
Number An ion's position vector is initially 7 = (2.4 m )î + (−7.4 m )3 + (6.5 m ). In unit-vector notation, what is its average velocity during the 6.7 s? )ĵ i 1 + = 9.1 m )i + (−9.9 m )ĵ + (−10 m )k, ) + (-10 m ), and 6.7 s later it is -0.37313433 Ĵ + MI -2.46268657 k Units m/s
Suppose the position vector for a particle is given as a function of time by r(t) = x(t)î + y(t)j, with x(t) = at + b and y(t) = ct + d, where a = 2.00 m/s, b = 1.15 m, c = 0.116 m/s2, and d = 1.08 m. (a) Calculate the average velocity during the time interval from t = 2.25 s to t= 3.80 s. m/s (b) Determine the velocity at t = 2.25 s. v = m/s Determine the speed at t = 2.25 s. m/s
A snowmobile is originally at the point with position vector 27.1 m at 95.0° counterclockwise from the x axis, moving with velocity 4.13 m/s at 40.0°. It moves with constant acceleration 15 m at 200 After 5.00 s have elapsed, find the following. (Express your answers in vector form.) (a) its velocity vector (b) its position vector
The acceleration of a particle moving only on a horizontal xy plane is given by a = 611 +71), where a is in meters per second- squared and t is in seconds. At t = 0, the position vector 7= (23.0m)i + (45.0m) locates the paticle, which then has the velocity vector 7 = (6.40m/s) + (2.60m/s)). At t - 3,40s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis?
A particle leaves the origin with an initial velocity v→=(2.33î) m/s and a constant acceleration a→=(-4.30î-2.65ĵ) m/s2. When the particle reaches its maximum x coordinate, what are (a) its velocity, (b) its position vector?

SEE MORE QUESTIONS