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An electron moves at 5.20 m/s at an angle of 36.8° above the horizontal. Three seconds later, its velocity is 7.18 m/s at an angle of 63.3° below the horizontal. What was the electron's average acceleration during these 3.00 seconds?

Guest Feb 25, 2018
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Acceleration is a change in velocity.

Let's break the electron's VELOCITY before and after the 3 seconds in to its x and y components.

vox = original x velocity  = 5.2 cos 36.8 = 4.164 m/s

voy =                                  5.2 sin 36.8 = 3.115

 

v1x = final x velocity      = 7.18 cos (-63.3) = 3.226 m/s

v1y =                                 7.18 sin (-63.3) = - 6.414

 

Now the CHANGE in velocities (x and y vectors)  is  

v1x-vox  and  v1y-voy

(3.226-4.164)  & (-6.414-3.115)    (these are x and y VECTORS so we cannot just add them together)

-0.938 m/s              -9.529          The RESULTANT change in velocity = sqrt (vx^2 +vy^2)

change in velocity  v delta= 9.575 m/s  @  -95.62 degrees 

This CHANGE in veocity happened over 3 seconds

Acceleration = change in velocity/change in time  = 9.575/3 = 3.192 m/sec^2  (absolute value)

   Note that the velocity change was negative for both x and y so the acceleration is - 3.192 m/sec^2

ElectricPavlov  Feb 25, 2018

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