Question #158621

A puck is moving on an air hockey table. Relative to an xy coordinate system at time t = 0 s, the xcomponents of the puck's initial velocity and acceleration are v0x = +1.2 m/s and ax = +6.6 m/s2. The ycomponents of the puck's initial velocity and acceleration are v0y = +6.5 m/s and ay = -3.3 m/s2. Find (a) the magnitude v and (b) the direction θ of the puck's velocity at a time of t = 0.50 s. Specify the direction relative to the +x axis


1
Expert's answer
2021-01-28T20:11:03-0500

The x componet of the velocity after t=0.5st = 0.5s is:


vx=v0x+axtvx=1.2m/s+6.6m/s20.5=4.5m/sv_x = v_{0x} + a_xt\\ v_x =1.2m/s + 6.6m/s^2\cdot 0.5 = 4.5m/s

The y componet of the velocity after t=0.5st = 0.5s is:


vy=v0y+aytvy=6.5m/s3.3m/s20.5=4.85m/sv_y = v_{0y} + a_yt\\ v_y =6.5m/s -3.3m/s^2\cdot 0.5 =4.85m/s

The magnitude of the velocity is then (by Phyphagorean's theorem):


v=vx2+vy2v=4.52+4.8526.62m/sv = \sqrt{v_{x}^2+v_{y}^2}\\ v = \sqrt{4.5^2+4.85^2} \approx 6.62m/s

The angle relative to the positive x axis is:


θ=arctan(vyvx)θ=arctan(4.854.5)47.1°\theta = \arctan\left(\dfrac{v_y}{v_x} \right)\\ \theta = \arctan\left(\dfrac{4.85}{4.5} \right) \approx 47.1\degree

Answer. a) 6.62m/s, b) 47.1 degree.


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