Example 1: Physics Relationships

A particle moves along a line so that at time $t$ its position is given by $s\left(t\right)=\frac{1}{2}{t}^3–\frac{7}{2}{t}^2+5t+20$.

i)                     What is the velocity of the particle when its acceleration is zero?

ii)                   Is the particle speeding up or slowing down at time t=4?

Step 1: Determine which equation, equations, or graphs you are given to start with.

The only equation you are given to start with is the position equation , $s\left(t\right)$.

$s\left(t\right)=\frac{1}{2}{t}^3–\frac{7}{2}{t}^2+5t+20$

Step 2: Determine which of your physics equations you will need to use to answer the question , and which physics equation you will need to use to find the time .

i)                     You need to know the velocity when the acceleration is zero.

ii)                   You need to know both the velocity and acceleration equations so you can compare their values at the given time .

i)                     Answer equation = velocity

Time equation = acceleration

ii)                   Answer equation = velocity and acceleration

Time equation = time given

Step 3: Take whatever derivatives you need to find your additional physics equations.

In this example you will need to find the derivative of the position equation to find the velocity equation, and then the derivative of that velocity equation to find the acceleration equation.

$s\left(t\right)=\frac{1}{2}{t}^3–\frac{7}{2}{t}^2+5t+20$

$s\prime \left(t\right)=v\left(t\right)=\frac{3}{2}{t}^2–7t+5$

$s\prime \prime \left(t\right)=v\prime \left(t\right)=a\left(t\right)=3t–7$

Step 4: Find any times that you need using the necessary physics equation.

Here you need to know “when the acceleration is zero.”

You will want to set your acceleration equal to zero and solve for that t-value , that time .

$a\left(t\right)=3t–7$

$3t–7=0$

$3t=7$

$t=\frac{7}{3}$

Step 5: Plug any times you have into the corresponding physics equation that is required by the question.

i)                     You will need to plug the $t=\frac{7}{3}$into the velocity equation to determine the velocity at that time .

ii)                   You will need to plug t=4 into both the velocity and acceleration equations.

i)                     $v\left(\frac{7}{3}\right)=\frac{3}{2}{\left(\frac{7}{3}\right)}^2–7\left(\frac{7}{3}\right)+5$

$v\left(\frac{7}{3}\right)=–\frac{19}{6}$

ii)                   $v\left(4\right)=\frac{3}{2}{\left(4\right)}^2–7\left(4\right)+5=1$

$a\left(4\right)=3\left(4\right)–7=5$

Final Results:

i)                     The velocity of the particle at the time the acceleration is zero would be $–\frac{19}{6}$.

If you are given units in the problem, make sure your answer includes the units.

ii)                   Since both velocity , $v\left(4\right)=1$,and acceleration , $a\left(4\right)=5$,are positive , the particle is speeding up .