Given a graph of an object's acceleration vs time. When t=0 the position and velocity of the object are both zero. The graph crosses the time axis at 3 and 5. At t=3 would the object be moving forward, backward or is it still?

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At #t=3# and #t=5# there is #0# acceleration. It means that at those two points in time. the object has a constant velocity.

We also know that at #t=0# both the position and the velocity were #0#. This mean the object was at rest which also means acceleration #a=0#.

Between #t=0# and #t=3#, the object began moving, i.e. velocity #v# and acceleration #a# increased.

This means that between #t=0# and #t=3# the acceleration would have to have been positive. As such, for acceleration to come back to #0# the velocity would have to become constant.

Therefore, the object is moving in the positive direction with a constant velocity at #t=3#.

The answer to your question of whether at #t=5# the object is moving backward is #"No"#.

The reason becomes apparent if we analyze what is actually happening.

We know that the acceleration curve crosses the time axis at three points, #t=0, 3, and 5#

Let's develop the equation of acceleration function:

#t(t-3)(t-5)=0#

#a(t)=t^3-8t^2+15t#

We now take the integral of the acceleration function to arrive at the velocity function:

#v(t)=inta(t)dt=int(t^3-8t^2+15t)dt=1/4t^4-8/3t^3+15/2t^2+c_1#

We can use our initial conditions to solve for #c_1#. At #t=0# velocity is #0#. If we plug this into the velocity function we see that #c_1=0#. Therefore,

#v(t)=1/4t^4-8/3t^3+15/2t^2#

We now take the integral of the velocity function to arrive at the position (distance) function:

#s(t)=1/20t^5-2/3t^4+5/2t^3+c_2#

Again, using initial conditions of #t=0# giving us #s=0# makes #c_2=0#. Therefore,

#s(t)=1/20t^5-2/3t^4+5/2t^3#

Now that we have all three functions, we can evaluate each at #t=3# and at #t=5#. We will see that both the velocity and position are positive at both #t# values.

The figure below shows all three functions on one coordinate system:

The red curve is the acceleration.

The purple curve is the velocity.

The blue curve is the position.

As you can see, for #t >0#, both the velocity and position curves all always above the #t#-axis which means they are both always positive.

As such, the object is always moving forward.