Question

A particle’s velocity is described by the function $v_x=\left(t^2-7t+10\right)\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$, where t is in s. a. At what times does the particle reach its turning points? b. What is the particle’s acceleration at each of the turning points?

Short answer

Part (a)

Particle reaches the turning point at $t=2s$and $t=5s$.

Part (b)

Acceleration of a particle at $t=2s$is $-3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.

Acceleration of a particle at$t=5s$is$3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.

Step-by-step solution

Given information

The velocity of the particle is given by the function$v_x=\left(t^2-7t+10\right)\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Part (a)

The turning point is a point of motion where a particle reverses its direction. It is a point where velocity instantly becomes zero and position is maximum or minimum.

At the turning point, $v_x=0$.

Substitute 0 for $v_x$into the given function.

$$\begin{gathered} 0=t^2-7t+10 \\ {t}^2-5t-2t+10=0 \\ t\left(t-5\right)-2\left(t-5\right)=0 \\ \left(t-5\right)\left(t-2\right)=0 \\ t=5ort=2 \end{gathered}$$

Therefore, the particle reaches the turning point at$t=5s$and$t=2s$.

Part (b)

Acceleration is given by $a_x=\frac{d{v}_x}{dt}$.

The acceleration of a particle is

$$\begin{gathered} a_x=\frac{d}{dt}\left(t^2-7t+10\right) \\ =2t-7 \end{gathered}$$

An acceleration of a particle at $t=2s$is

$$\begin{gathered} a_x=2\left(2\right)-7 \\ =4-7 \\ =-3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

An acceleration of a particle at $t=5s$is

$$\begin{gathered} a_x=2\left(5\right)-7 \\ =10-7 \\ =3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Therefore, an acceleration of a particle at $t=2s$and $t=5s$is $-3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$and $3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$respectively.