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

#1**+1 **

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