Where on the graph is the acceleration a>0, a<0, and a=0?

Question

The graph of the position of an oscillating object as a function of time is shown. http://session.masteringphysics.com/problemAsset/1010963/14/MHM_va.jpg

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Where on the graph is the acceleration a > 0?

A: A to B

B: A to C

C: C to D

D: C to E

E: B to D

F: A to B and D to E

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Where on the graph is the acceleration a < 0?

A: A to B

B: A to C

C: C to D

D: C to E

E: B to D

F: A to B and D to E

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Where on the graph is the acceleration a = 0?

A: A only

B: B only

C: C only

D: D only

E: E only

F: A and C

G: A and C and E

H: B and D

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Choose the most complete answer. For example, if the answer "B to D" were correct, then "B to C" would technically also be correct. Also please try to explain your reasoning.

Answer 1

You know in mechanics that acceleration a(t) the second derivative of x(t)

a(t) = x&quot;(t)

You also know in calculus that the sign of the second derivative of a function in an interval shows the direction of the concavity of the graph in that interval.

x&quot;(t) &gt; 0 &lt;--&gt; the concavity faces toward positive direction

x&quot;(t) &lt; 0 &lt;--&gt; the concavity faces toward negative direction

x&quot;(t) = 0 at inflexion points, where concavity changes its orientation.

You can apply these laws to find the answers to your questions

Answer 2

1) D

2) B

3) G