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Classical mechanics is a fundamental part of physics that deals with the motion of macroscopic objects in the presence of forces. It is based on the laws of motion proposed by Sir Isaac Newton in the 17th century, which form the foundation of classical mechanics.
In classical mechanics, we use various quantities to describe the behavior of objects, including position, velocity, acceleration, force, work, energy, momentum, and angular momentum. These quantities are related through mathematical equations and principles that describe how objects move and interact with each other.
For example, the work-energy principle states that the work done on an object is equal to the change in its kinetic energy, while the law of conservation of momentum states that the total momentum of a closed system remains constant unless acted upon by an external force.
Classical mechanics is still widely used today in many fields of science and engineering, from astronomy to robotics, and it provides a solid foundation for understanding the behavior of objects at the macroscopic level.
A runner travels along the $\mathrm{x}$-axis, and at time $t=0$ is at the origin. The $x$-component of the runner’s position with respect to the origin is given by: $$ x(t)=b t^2 $$ where $b$ is a positive constant. (Part a) What are the units of the constant $b$ ? Express your answer in terms of $\mathrm{m}$ for meter and $\mathrm{s}$ for seconds. (Part b) Find $v(t)$, the $x$-component of the runner’s velocity as a function of time.
a) The units of the constant $b$ can be obtained by looking at the units of $x(t)$. We have $x(t)=[b][t^2]$, where $[b]$ denotes the units of $b$ and $[t]$ denotes the units of $t$. Since $x(t)$ has units of length (meters), we have $[b]=\mathrm{m}/\mathrm{s}^2$.
b) The velocity of the runner is given by the derivative of the position function with respect to time:
$v(t)=\frac{d x}{d t}=\frac{d}{d t}\left(b t^2\right)=2 b t$.
Therefore, the $x$-component of the runner’s velocity as a function of time is $v(t) = 2bt$.
(Part c) Find $a(t)$, the $x$-component of the runner’s acceleration as a function of time.
c) The acceleration of the runner is given by the second derivative of the position function with respect to time:
$a(t)=\frac{d^2 x}{d t^2}=\frac{d}{d t}(2 b t)=2 b$
Therefore, the $x$-component of the runner’s acceleration as a function of time is constant and given by $a(t) = 2b$.
At time $t=0$, a car moving along the $+x$-axis passes through $x=0$ with a constant velocity of magnitude $v_0$. At some time later, $t_1$, it starts to slow down. The acceleration of the car as a function of time is given by:
$$
a(t)= \begin{cases}0 & 0 \leq t \leq t_1 \ -c\left(t-t_1\right) & t_1<t \leq t_2\end{cases}
$$
where $c$ is a positive constants in $\mathrm{Sl}$ units, and $t_1<t \leq t_2$ is the given time interval for which the car is slowing down. The goal of the problem is to find the car’s position as a function of time between $t_1<t<t_2$. Express your answer in terms of $\mathrm{v}{-} 0$ for $v_0$, t 1 for $t_1, \mathrm{t} 2$ for $t_2$, and $c$ as needed.
(Part a). What is $v(t)$, the velocity of the car as a function of time during the time interval $0 \leq t \leq t_1$ ?
Since the acceleration of the car is zero during the time interval $0 \leq t \leq t_1$, the car’s velocity remains constant during this time interval. Therefore, the velocity of the car as a function of time during the time interval $0 \leq t \leq t_1$ is given by:
$$v(t) = v_0 \qquad \text{for } 0 \leq t \leq t_1$$
where $v_0$ is the constant velocity of the car at $t=0$.
