Question

The velocity $\overrightarrow{\mathbf{v}}$of a particle moving in the xy plane is given by $\overrightarrow{\mathbf{v}}\mathbf{=}\left(6.0t-4.0t^2\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{-}\left(8.0\right)\stackrel{\mathbf{⏜}}{\mathbf{j}}$, with $\overrightarrow{\mathbf{v}}$in meters per second and t(>0) in seconds. (a) What is the acceleration when t=3.0 s ? (b) When (if ever) is the acceleration zero? (c) When (if ever) is the velocity zero? (d) When (if ever) does the speed equal 10 m/s ?

Short answer

(a) The acceleration when time t=3.0 s is$\left(-18\mathrm{m}/{\mathrm{s}}^2\right)\stackrel{⏜}{\mathrm{i}}$

(b) Acceleration is zero when t=0.75s

(c) Velocity can never be zero in the given situation.

(d) Speed is zero when t=2.2 s

Step-by-step solution

Given information

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

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

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

To understand the concept

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,

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

(a) To find the acceleration when t=3.0 s

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

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

Now, using equation (i),

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

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

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

(b) To find the time when acceleration is zero

It is given that,acceleration,

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

(c) To find the zero velocity

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

(d) To find the time when speed equal 10 m/s

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

Therefore, the speed of particle is,

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

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,

$\left(6.0t-4.0t^2\right)=\pm 6$

Thus,

t=2.2 s