The magnitude of velocity of a car moving in a vacant parking lot is [tex]\fbox{\begin\\7.467\,{\text{m/s}}\end{minispace}}[/tex] and direction of velocity is [tex]\fbox{\begin\\58.984^\circ\end{minispace}}[/tex] with the positive direction of [tex]X[/tex] 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
[tex]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[/tex].
Concept:
From the given expression of velocity it can be observed that the velocity of the car has its components in [tex]X[/tex] and [tex]Y[/tex] direction.
The [tex]X[/tex] component of velocity is:
[tex]\fbox{\begin\\{V_x}=5 - 0.0180\,{t^2}\end{minispace}}[/tex]
The [tex]Y[/tex] component of velocity is:
[tex]\fbox{\begin\\{V_y}=2 + 0.55\,t\end{minispace}}[/tex]
At [tex]t=8.00\,{\text{s}}[/tex].
The [tex]X[/tex] component of velocity is:
[tex]\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}[/tex]
Similarly the [tex]Y[/tex] component of velocity is:
[tex]\begin{aligned} {V_y}&=2 + 0.55\left( {8.00} \right)\\ &=6.4\text{ m/s}} \\ \end{aligned}[/tex]
The magnitude of velocity of car is:
[tex]\fbox{\begin\bar V=\sqrt {V_x^2 + V_y^2}\end{minispace}}[/tex]
Substitute [tex]3.848\,{\text{m/s}}[/tex] for [tex]{V_x}[/tex] and [tex]6.4\,{\text{m/s}}[/tex] for [tex]{V_y}[/tex] in the above expression .
[tex]\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}[/tex]
Let [tex]\theta[/tex] is the angle made by the resultant velocity with the positive direction of [tex]X[/tex] axis.
The direction of velocity at time [tex]t = 8.00\,{\text{s}}[/tex] is calculated by:
[tex]\fbox{\begin\\\theta={\tan ^{ - 1}}\left( {\frac{{{V_y}}}{{{V_x}}}} \right)\end{minispace}}[/tex]
Substitute [tex]3.848\,{\text{m/s}}[/tex] for [tex]{V_x}[/tex] and [tex]6.4\,{\text{m/s}}[/tex] for [tex]{V_y}[/tex] in the above expression .
[tex]\begin{aligned} \theta&={\tan ^{ - 1}}\left( {\frac{{6.4\,{\text{m/s}}}}{{3.848\,{\text{m/s}}}}} \right) \\ &=58.984^\circ \\ \end{aligned}[/tex]
Thus, the magnitude of velocity of a car moving in a vacant parking lot is [tex]\fbox{\begin\\7.467\,{\text{m/s}}\end{minispace}}[/tex] and direction of velocity is [tex]\fbox{\begin\\58.984^\circ\end{minispace}}[/tex] with the positive direction of [tex]X[/tex] axis.
Learn more:
1. Conservation of energy brainly.com/question/3943029
2. A ball falling under the acceleration due to gravity brainly.com/question/10934170
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.
