An electron has a constant acceleration of


An electron has a constant acceleration of +3.2 m/s^2. At a certain instant its velocity is 9.6 m/s. What is its velocity (a) 2.5 s earlier and (b) 2.5 s later?

Solution:

To solve this question, we will use the following equation:

v=v_0+at

(where v is final velocity, v_0 is initial velocity, a is acceleration, and t is time.)

 
a) To find the velocity before 2.5 seconds, we will set t=-2.5 s. Thus, we have:

v=(9.6 m/s)+(3.2 m/s^2)(-2.5 s)

v=1.6 m/s

 

b) To find the speed after 2.5 seconds, we will set t=+2.5 s. Thus, we have:

v=(9.6 m/s)+(3.2 m/s^2)(2.5 s)=18 m/s

 

This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 27.

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