(c) Find the acceleration vector ā(t). What direction does the acceleration vector point in? Hint: The acceleration is the time-derivative of the velocity.

College Physics
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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)...
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An object is moving in a circle of radius 10 cm centered on the origin in the xy-plane at a constant speed of 5 m/s.
Transcribed Image Text:An object is moving in a circle of radius 10 cm centered on the origin in the xy-plane at a constant speed of 5 m/s.
(c) Find the acceleration vector \(\vec{a}(t)\). What direction does the acceleration vector point in? Hint: The acceleration is the time-derivative of the velocity.
Transcribed Image Text:(c) Find the acceleration vector \(\vec{a}(t)\). What direction does the acceleration vector point in? Hint: The acceleration is the time-derivative of the velocity.
Expert Solution
Step 1

Given: Radius of the circle is r = 10 cm

centered at origin (0, 0) and the linear velocity of the particle is 

v = 5 m/s Representing the circular motion in vector form:

                                               Physics homework question answer, step 1, image 1

c)

From the figure position vector of the particle at any time t  is 

r(t) = xi^ +yj^                                                                            

Now the angular velocity of the particle is given by:

ω = vr=5 m/s0.1 m    = 50 rad/s

Angle of the particle at time twill be

θ = ωt

hence the position coordinates of the particle at this moment will be:

x = r cos(ωt)y = r sin(ωt)

then by equation 1 the position vector will be:

r(t) = (r cos(ωt)i^ +(r sin(ωt))j^      =0.1{cos(50t)i^ + sin(50t)j^}

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