Question

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?

Answer 1

Moving forward

Explanation:

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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.