The position versus time graph above shows the position of a mouse in meters, between the times t = 0 s and t = 5.00 s.

The mouse moves as described by the equation above.

Variables

We use the convention that quantities toward the right are positive, and quantities toward the left are negative.

What is the strategy?

1. The velocity is the time rate of change of the position. Differentiate the position expression to find an expression for the velocity as a function of time.
2. Since the velocity function is continuous, it can change sign from negative to positive, or vice versa, only at times when it equals zero. Solve for the times when the velocity is zero. (This can be checked by looking for points on the position-time graph where the slope is horizontal.)
3. Find the sign (direction) of the velocity during all the relevant time intervals, by evaluating the velocity function at a time value within an interval. (This can be checked by noting the sign of the slope of the position graph during each interval.)
4. Identify the interval(s) during which the velocity is negative.
5. The acceleration is the time rate of change of the velocity. Differentiate the velocity function to find an expression for the acceleration as a function of time.
6. Solve for the time(s) at which the acceleration is zero.

Physics principles and equations

The function giving the velocity as a function of time is the derivative of the position function.

The function giving the acceleration as a function of time is the derivative of the velocity function, or the second derivative of the position function with respect to time.

Mathematics principles

The derivative of a sum is the sum of the individual derivatives.

The derivative of Kt ⁿ is nKt ⁿ⁻¹.

Step-by-step solution

We have determined the time at which the acceleration is zero. We can check this result using the position graph. The graph is concave down for times earlier than this, which tells us that the second derivative is negative − in other words, the acceleration is negative. The graph is concave up for times later than this − so the acceleration is positive. This time represents a point of inflection, when the acceleration is zero.