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avatar+179 

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motions

 = 2 sin(π t) + 2 cos(π t),

 where t is measured in seconds. (Round your answers to two decimal places.)

 

Find the average velocity during each time period.

 

(i)    [1, 2] (1 to 2 seconds)
  ____cm/s 

(ii)    [1, 1.1] 
____ cm/s 

(iii)    [1, 1.01] 
 ____cm/s 

(iv)    [1, 1.001] 
 ____cm/s 


(b) Estimate the instantaneous velocity of the particle when t = 1. 
____cm/s

 Jan 29, 2019
 #1
avatar+95866 
+1

2[  sin (pi *t  ) + cos (pi * t)

 

From 1 to 2 seconds we have

 

[  2 sin (2pi) + 2cos  (2 pi)  - 2sin (pi) - 2cos(pi)] / (2 - 1) =

 

[ 2  - (-2) ] / 1      =     4 cm /sec

 

 

From 1 to 1.1 seconds

 

[  2 sin (1.1pi) + 2cos  (1.1 pi)  - 2sin (pi) - 2cos(pi)] / (1.1 - 1) ≈  -5.20 cm/sec

 

From 1 to 1.01 seconds

 

[  2 sin (1.01*pi) + 2cos  (1.01* pi)  - 2sin (pi) - 2cos(pi)] / (1.01 - 1) ≈ -6.18 cm/sec

 

From  1 to 1.001 seconds

 

[  2 sin (1.001*pi) + 2cos  (1.001* pi)  - 2sin (pi) - 2cos(pi)] / (1.001 - 1)  ≈ - 6.27 cm /sec

 

It appears that  the instantaneous velocity at 1 sec   =  -2pi  cm/sec  ≈  -6.28 cm/sec

 

BTW....we can verify that this is true using some Calculus   !!!

 

 

 

 

cool cool cool

 Jan 29, 2019
 #2
avatar+179 
+1

oh wow that's a lot. thank you

Ruublrr  Jan 29, 2019
 #3
avatar+95866 
0

You're welcome.....!!!!

 

 

cool cool cool

CPhill  Jan 29, 2019

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