Question

6. 2762	Given a set of time and velocity data: t=0:10 v=[0 0.5 1.1 1.9 3.1 5.4 8.1 10.1 13.1 15 19] Use MATLAB's diff command to determine the acceleration over each interval. Use the forward difference method and interp1 to calculate the acceleration at t=6.7 seconds

Part	Description	Answer	Save	Status
A.	Use the forward difference method and interp1 to calculate the acceleration at t=6.7 seconds (no units)

Answer 1

>>t=0:10;
>>v=[0 0.5 1.1 1.9 3.1 5.4 8.1 10.1 13.1 15 19];
>>a = diff(v)./diff(t); %take difference of velocity and divide each element by difference time.

>>a=

0.5000 0.6000 0.8000 1.2000 2.3000 2.7000 2.0000 3.0000 1.9000 4.0000

A)

>>interpV = interp1(t,v,6.7)

>>interpV  = 9.5000

Now we need to use this velocity at 6.7s and find acceleration at 6.7s.

Formula for forward difference is (f(x+d) - f(x)) / (x+d -x). Here x=6s, d=0.7s and f(x)= v(t)=velocity @ time = t

= (v(6+0.7)-v(6)) / (6.7-6)

>>interpA = (interpV - v(7)) / (6.7 - t(7))

>>interpA = 2.0

Why t(7) and v(7) is because to find the diff we need take velocity and time which is at t=6s. And t=6s is at index 7. Verify the index by counting in t vector.