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Medicine: Physics: Linear Motion with constant acceleration

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### Linear Motion with constant acceleration

Kinematics of a particle

3.1 Introduction

Kinematics is the study of particle motion without reference to mass or force. In some ways, studying kinematics is rather artificial: in almost all realistic situations, the motion would have been produced by forces and the problem can only be solved by investigating the equations of motion appropriate to the forces acting. The study of motion produced by forces is called Dynamics. Note that we deal with particles, which, by definition, are point-like; they can have mass (though that is not needed in kinematics) but they have no internal structure, so they cannot, for example, The example of projectile flight is important, historically and in terms of applications. From our point of view, it is a first stab at tackling equations of motion, which is fundamental to all theoretical physics courses.

3.2 Key concepts

Differentiation of a vector is the same as differentiation of its Cartesian components.

Definitions of speed, velocity and acceleration

Formulae for particles moving with constant acceleration in one dimension.

3.3 Notation

Motion on a line:

In one (spatial) dimension, the variables are time, position or distance or displacement from a fixed point, speed or velocity, and acceleration. We make a distinction between speed and velocity even in one dimension: velocity may be positive or negative, corresponding to the particle moving (say) to the right or left; speed is the magnitude of velocity and is therefore always positive or zero. Acceleration can also be either positive or negative. We denote time by $t$, position by $x$, velocity by $u$ or $v$ and acceleration by $a$. Sometimes, displacement from the original position of the particle is denoted by $s$. We might write, for example,$x(t)$ to emphasise that x is a function of time. Velocity, by definition, is rate of change of position, so$v=\frac{dx}{dt}=\dot{x}$

The overdot always denotes differentiation with respect to time.Acceleration, by definition, is rate of chance of velocity, so

$a=\frac{dv}{dt}=\dot{v}\frac{d^2x}{dt^2}=\ddot{x}$

3.4 Constant acceleration on a line

We can obtain standard results for constant acceleration by (as is often the case) writing down the definitions and integrating the resulting differential equations. We have

$a=\ddot{x}$

where $a$ is constant, so integrating once gives$\dot{x}=v=at+u$

where $u$ is a constant of integration corresponding to the velocity at $t=0$. Integrating again gives

$x=\frac{1}{2}at^2+ut+x_0$

where $x_0$ is a constant of integration corresponding to the position at $t=0$ . Sometimes, this is written as

$s=\frac{1}{2}at^2+ut$

where $s$ is displacement from the initial position. It is worth checking dimensions at each stage as a quality control check. The dimension of $s$ is length $(L)$, the dimension of $u$ is $L/T$ and the dimension of a is $L/T^2$ substituting these into this last equation reveals it is dimensionally consistent. Equation (2) gives distance as a function of time. We can find distance as a function of velocity by using (1) to eliminate time from (2):which gives

$2as=v^2-u^2$

Examples of application:

A car starts from rest on a straight road, and moves with a constant acceleration of $2ms^{-2}$

1. Calculate its velocity at a time $t=5s$
2. Calculate the distance traveled after $10s$
3. Given that the car decelerates to rest from a speed $u$, find $u$ if it travels a distance of $5m$  with an acceleration of$-2ms^{-2}$ to get to rest from $u$.

Solution:

1. by using the formula $v=at+u$ since the car starts from rest $u=0m/s$

$v=(2s)*(5ms)^2+0ms^{-1}$

$v=10ms^{-1}$

2.      From $s=\frac{1}{2}+ut$

$S=0.5(2ms^{-2})(10s)^2$

$S=100m$

3.      From $2as=v^2-u^2$

$-u^2=2(-2m/s^{-2})(5m)$

U=$2\sqrt{5}m/s$

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