Should #16 be option c (twice as much as the Lanchester)?
v = sqroot(2 * 4a* x) = sqrt (4) *sqrt (2ax) = 2 v lanchester

Michael, You are correct, thanks for the catch. The files have been updated.  John

For Question 24, shouldn't the acceleration be 2 m/s/s? The answer key has it as 4 m/s/s.  Thank you.

Joan, you need to use the second kinematics equation where x initial and v initial are equal to zero.  x = 1/2 at^2.
Rearrange to solve for a = 2x/t^2. Then substitute in values for x and t at any of the points, and each point will give a = 4 m/s^2.  John

Thank you, John.  That makes sense.  Now this is my question.  What cue was in that problem that should have directed me to use the second kinematics equation, not the first one.  If I selected to use the first one, then I would have used the formula:  a= (delta velocity/delta time).  This equation would have brought me to 2 m/s/s.  However, if I choose to use the second one, then I would have arrived at the answer as 4 m/s/s.

Joan, that means you found the average velocity over the 5 s which is delta x over delta t = 50m/5s = 10 m/s. Then you divided that by 2 s and got a = 2 m/s^2. 

The problem is that is the average velocity. Acceleration is defined as the (final velocity minus the initial velocity)/ delta time. We don't have the final velocity - which is why we used the second kinematics equation. 

The actual final velocity can then be found using the first or third kinematics equation:

v final = v zero + at = 0 +(4 m/s^2)(5s) = 20 m/s

or 

v final = sqr root (initial velocity squared + 2 a delta x) = sqrt((0 + 2(4 m/s^2)(50m)) = 20 m/s. 

And to complete the analysis: 

v avg = (v final + v original)/2 = (20 +0)/2 m/s = 10 m/s.

This thread will be helpful to many teachers and students, thanks for the questions!

John

Thanks again!