Question

A particle’s velocity is given by the function vx = 12.0 m/s2sin1pt2,where t is in s. a. What is the first time after t = 0 s when the particle reaches a turning point? b. What is the particle’s acceleration at that time?

Short answer

Part (a)

The first time after $t=0s$when a particle reaches a turning point is $t=1s$.

Part (b)

Acceleration of the particle at the turning point islocalid="1649476593269" $6.3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.

Step-by-step solution

Given information

The velocity of the particle is given by a function$v_x=\left(2.0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)\mathrm{sin\pi }t$.

Part (a)

The turning point is the point in motion where a particle reverses its direction.

It is a point where velocity is instantaneously zero and position is maximum or minimum.

The velocity is zero when

$$\begin{gathered} 0=2\mathrm{sin\pi }t \\ \mathrm{sin\pi }t=0 \\ t\Rightarrow 0s,1s,... \end{gathered}$$

Therefore, the particle reaches its turning point after the first time at$t=1s$.

Part (b)

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

Differentiate given expression with respect to $t$.

localid="1649476489349" $$\begin{gathered} a_x=\frac{d}{dt}\left(2\mathrm{sin\pi }t\right) \\ =2\mathrm{\pi cos\pi }t \end{gathered}$$

The acceleration of the particle at $t=1s$is

localid="1649476559659" $$\begin{gathered} a_x\left(att=1s\right)=2\mathrm{\pi cos\pi }\left(1\right) \\ =6.3\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{s}}^2$}\right. \end{gathered}$$

Therefore, the acceleration of the particle at turning point islocalid="1649476571694" $6.3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.