Solve Physics Q: Acceleration & Velocity of Object

[Kennedy]

An object, moving in a straight line, has a velocity of 5.0 m/s at time t = 0. From t = 0 to t = 5 s its acceleration is 2.5 m/s2 , while from t = 5 to t = 11 s its acceleration is - 0.1 m/s2 . Over the time interval t = 5 s to t = 11 s its average velocity is: (a) 15.0 m/s (b) 12.5 m/s (c) 17.2 m/s (d) 14.5 m/s (e) 8.8 m/s

I tried the question by finding the distance covered by the particle after the first time interval, then the second. I subtracted my first distance from the second to have the total distance traveled by the object between 5 and 11 seconds. Then, I divided by 6 for the change in time. My answer was 14.5 m/s, but apparently that's not the right answer. It should be 17.2 m/s according to my answer key, but I have no idea how they arrived at that answer. Can someone please work out the solution to this problem for me, please? I'm going based off of that the average velocity of an object is d/t

 

[phyzguy]

Your method is OK, but you must have made a mistake. Please show your calculations. At t=5s, the velocity is 17.5 m/s. At t=11s, it is 16.9 m/s. How could the average possibly be 14.5 m/s, smaller than both the velocity at the beginning and at the end?

 

[Kennedy]

I had my distance covered in the first time interval to be 56.25 m (from 0 to 5), and my second distance to be 103.56 m (from 5 to 11, starting with an initial speed of 17.5 m/s). So, then the distance covered from 5 to 11 would be 103.56 m - 56.25 m, right? ...and then that answer divided by 6 (time interval)?

 

[Kennedy]

Oh, I think I figured it out by talking it through. The answer should simply be 103.56/6... which yields 17.2 m/s! THANKS.