Question

Question #bfeba

Answer 1

After `0.46` seconds.

Explanation:

So we want to find when the velocity is `0`. We know that the rate of change of displacement is velocity. Therefore, we find the derivative with resect to `s` of the displacement function.

`s' = 2e^(2t) -5`

This will equal `0` when:

`0 = 2e^(2t) - 5`

`5/2 = e^(2t)`

`ln(5/2) = ln(e^(2t))`

`ln(5/2) = 2t`

`1/2ln(5/2) = t`

Using a calculator, we get this will occur at approximately `0.46` seconds. We could also solve this problem graphically. Whenever a horizontal tangent may be drawn on the graph of the displacement-time graph, the object will be at rest. We can see from the following graph that the only time this occurs is at `t= 0.46`.

Hopefully this helps!