91Ó°ÊÓ

It is given that, velocity$\stackrel{→}{v}$of particle moving in xy plane is,

$\stackrel{→}{\mathrm{v}}=\left(6.0\mathrm{t}-4.0{\mathrm{t}}^2\right)\stackrel{Áåœ}{\mathrm{i}}+8.0\stackrel{Áåœ}{\mathrm{j}}$

Where,$\stackrel{→}{v}$in m/s and t > 0s.

This problem deals with a simple algebraic operation that involves calculation of the time, acceleration and the velocity at given time.By differentiating the given function of velocity, the acceleration of the given object can be found. This acceleration and velocity function can be used to answer the above questions.

Formula:

The acceleration in general can be written as,

$\stackrel{→}{a}=\left(\frac{d\stackrel{→}{v}}{dt}\right)$

Velocity $\stackrel{→}{v}$of particle moving in xy plane is,

$\stackrel{→}{\mathrm{v}}=\left(6.0\mathrm{t}-4.0{\mathrm{t}}^2\right)\stackrel{Áåœ}{\mathrm{i}}+8.0\stackrel{Áåœ}{\mathrm{j}}$

Now, using equation (i),

$$\begin{gathered} \stackrel{→}{\mathrm{a}}=\left(\frac{\mathrm{d}\left(6.0\mathrm{t}-4.0{\mathrm{t}}^2\right)\stackrel{Áåœ}{\mathrm{i}}+8.0\stackrel{Áåœ}{\mathrm{j}}}{\mathrm{dt}}\right) \\ =\left(6.0-8.0\mathrm{t}\right)\stackrel{Áåœ}{\mathrm{i}} \end{gathered}$$

The acceleration role="math" localid="1656476502415" $\stackrel{Áåœ}{a}$at t=3.0s is,

$$\begin{gathered} \stackrel{→}{a}=\left(6.0-8.0×3\right)\stackrel{Áåœ}{i} \\ \stackrel{→}{a}=\left(-18\mathrm{m}/{\mathrm{s}}^2\right)\stackrel{Áåœ}{\mathrm{i}} \end{gathered}$$

It is given that,acceleration,

$$\begin{gathered} \stackrel{→}{a}=\left(6.0-8.0t\right)\stackrel{Áåœ}{i} \\ t=\frac{6.0}{8.0} \\ =0.75s \end{gathered}$$

Velocity $\stackrel{→}{v}=\left(6.0t-4.0t^2\right)\stackrel{Áåœ}{i}+8.0\stackrel{Áåœ}{j}$can never be zero, because the y component of velocitylocalid="1656477090473" $\stackrel{→}{v_v}=8.0\stackrel{Áåœ}{j}$cannot be zero.

It is given that, velocity$\stackrel{→}{v}=\left(6.0t-4.0t^2\right)\stackrel{Áåœ}{i}+8.0\stackrel{Áåœ}{j}$

Therefore, the speed of particle is,

$$\begin{gathered} v=\left|\stackrel{→}{v}\right|=\sqrt{\left(6.0t-4.0t^2\right)\stackrel{Áåœ}{i}+8.0\stackrel{Áåœ}{j}} \\ =10m/s \end{gathered}$$

Solving above equation for t,

$$\begin{gathered} {\left(6.0t-4.0t^2\right)}^2+64=100 \\ {\left(6.0t-4.0t^2\right)}^2+64=36 \end{gathered}$$

Taking square roots of both the sides,

$\left(6.0t-4.0t^2\right)=Â\pm 6$

Thus,

t=2.2 s