The following equations give the velocityof a particle in four situations:$\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathit{v}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathit{v}\mathbf{=}\mathbf{4}{\mathit{t}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{t}\mathbf{-}\mathbf{6}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathit{v}\mathbf{=}\mathbf{3}\mathit{t}\mathbf{-}\mathbf{4}\mathbf{}\mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathit{v}\mathbf{=}\mathbf{5}{\mathit{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}$.To which of these situations do the equations of Table 2-1 apply?

Table 2.1 with equations of motion with constant acceleration.

The problem uses the simple concept of acceleration which is the rate of change of velocity with time. The general formula for acceleration can be used to find the situations to which the equations in table 2.1 can be applied.

Formula:

Acceleration (a)$=\frac{dv}{dt}$

The equations of table 2.1 only apply to the situations in which acceleration is constant.

Acceleration can be defined as,

Acceleration (a)$=\frac{dv}{dt}$

(a) The given velocity fuction is ,v = 3.so,

$$\begin{gathered} a=\frac{d\left(3\right)}{dt} \\ =0 \end{gathered}$$

Acceleration is constant in this situation.

Therefore,the equations of Table 2.1 can be applied to (a).

role="math" localid="1656997781414" $v=4t^2+2t-6$

So

role="math" localid="1656997790862" $$\begin{gathered} v=\frac{d\left(4t^2+2t-6\right)}{dt} \\ =8t+2 \end{gathered}$$

Acceleration is not constant in this situation.

Therefore,the equations of table 2.1 cannot be applied to (b).

(c)So,

$$\begin{gathered} a=\frac{d(3t-4)}{dt} \\ =3 \end{gathered}$$

Acceleration is constant in this situation.

Therefore,the equations of Table 2.1 can be applied to (c).