Question

A remote controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v= [5.00 m/s – (0.0180 m/s2)t^2 ]i+[2.00 m/s + (0.550 m/s2)t ]j . What are the magnitude and direction of the velocity of the car at t= 8.00 s?

Answer 1

The magnitude of velocity of a car moving in a vacant parking lot is $\fbox{\begin\\7.467\,{\text{m/s}}\end{minispace}}$ and direction of velocity is $\fbox{\begin\\58.984^\circ\end{minispace}}$ with the positive direction of $X$ axis.

Further explanation:  

Velocity is the rate of change of position of a body with respect to time. The velocity of car moving in a vacant parking lot is a function of time. The magnitude of the velocity of the car will change as the time changes but the direction may or may not change with respect to time.

Given :

The velocity of car as a function of time is given by  

$V=\left[ {5.00\,{\text{m/s}}-\left( {0.0180\,{\text{m/}}{{\text{s}}^{\text{3}}}} \right){t^2}} \right]\hat i+\left[ {2.00\,{\text{m/s +}}\left( {0.550\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)t} \right]\hat j$.

Concept:  

From the given expression of velocity it can be observed that the velocity of the car has its components in $X$ and $Y$ direction.  

The $X$ component of velocity is:  

$\fbox{\begin\\{V_x}=5 - 0.0180\,{t^2}\end{minispace}}$  

The $Y$ component of velocity is:  

$\fbox{\begin\\{V_y}=2 + 0.55\,t\end{minispace}}$  

At $t=8.00\,{\text{s}}$.  

The $X$ component of velocity is:  

$\begin{aligned} {V_x}&=5 - 0.0180(8)^{2}\\&=3.848{\mkern 1mu} \text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$  

Similarly the $Y$ component of velocity is:  

$\begin{aligned} {V_y}&=2 + 0.55\left( {8.00} \right)\\ &=6.4\text{ m/s}} \\ \end{aligned}$  

The magnitude of velocity of car is:  

$\fbox{\begin\bar V=\sqrt {V_x^2 + V_y^2}\end{minispace}}$  

Substitute $3.848\,{\text{m/s}}$ for ${V_x}$ and $6.4\,{\text{m/s}}$ for ${V_y}$ in the above expression .

$\begin{aligned} \bar V&=\sqrt {{{\left( {3.848\,{\text{m/s}}} \right)}^2} + {{\left( {6.4\,{\text{m/s}}} \right)}^2}} \\ &=7.467\,{\text{m/s}} \\ \end{aligned}$  

Let $\theta$ is the angle made by the resultant velocity with the positive direction of $X$ axis.  

The direction of velocity at time $t = 8.00\,{\text{s}}$ is calculated by:  

$\fbox{\begin\\\theta={\tan ^{ - 1}}\left( {\frac{{{V_y}}}{{{V_x}}}} \right)\end{minispace}}$

Substitute $3.848\,{\text{m/s}}$ for ${V_x}$ and $6.4\,{\text{m/s}}$ for ${V_y}$ in the above expression .

$\begin{aligned} \theta&={\tan ^{ - 1}}\left( {\frac{{6.4\,{\text{m/s}}}}{{3.848\,{\text{m/s}}}}} \right) \\ &=58.984^\circ \\ \end{aligned}$  

Thus, the magnitude of velocity of a car moving in a vacant parking lot is $\fbox{\begin\\7.467\,{\text{m/s}}\end{minispace}}$ and direction of velocity is $\fbox{\begin\\58.984^\circ\end{minispace}}$ with the positive direction of $X$ axis.  

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3. The motion of a body under friction brainly.com/question/4033012

Answer Details:  

Grade: College  

Subject: Physics  

Chapter: Kinematics  

Keywords:  

Remote controlled car, vacant parking lot, function, time, V, v, 5.00 m/s, 5 m/s, 5 meter per second, (0.0180 m/s2 )t^2 I+ (0.550 m/s2)t j, magnitude, direction, t=8.00 s, 7.467 m/s, 58.984 degree, 2.00 m/s, 2 m/s.

Answer 2

The magnitude and direction of the velocity of the car at t = 8.00 s is v = 7,468 m / s and the vector direction = θ = 58,984°

Further explanation

In the coordinates of the two-dimensional plane (xy) there is a position vector of a point

You can write P as:

P = xi + yi

where i denotes the unit vector in the x and y directions

While the magnitude of the vector:

$\large{\boxed{\bold{P~=~\sqrt{x^2~+~y^2}}}}$

and its direction

$\large{\boxed{\bold{\theta~=~arc~tan\frac{y}{x}}}}$

Likewise with the instantaneous speed which is a derivative of the position function of time

$V =\frac{dP}{dt}$

$V ~=~\frac{dx}{dt}i~i~+\frac{dy}{dt}j$

V = vx i + vy j

The velocity of the car as a function of time is given by v = [5.00 m / s - (0.0180 m / s²) t²] i + [2.00 m / s + (0.550 m / s²) t] j so that:

vx = 5 - 0.0180t² i

vy = 2 + 0.55t j

So the instantaneous velocity at t = 8 s becomes:

vx = 5-0.0180.8²

vx = 3,848 m / s

vy = 2+ 0.55.8

vy = 6.4

So the velocity position vector becomes:

v = 3,848 i + 6.4 y

and the magnitude becomes:

$v=\sqrt{3.848^2+6.4^2}$

v = 7,468 m / s

While the velocity vector direction:

$\theta~=~=arc~tan\frac{vy}{vx}$

$\theta~=~arc~tan~\frac{6.4}{3.848}$

θ = 58,984

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Keywords: velocity, vector, two-dimensional coordinates,