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igor_vitrenko [27]
3 years ago
14

A flea walking along a ruler moves from the 45 cm mark to the 27 cm mark. It does this in 3 seconds. What is the speed? What is

the velocity?
(Define increasing numbers to be the positive direction and decreasing numbers to be the negative direction.)
Physics
1 answer:
Alekssandra [29.7K]3 years ago
8 0

Answer:

Speed= 6cm/s and velocity= 6cm/s in the negative direction

Explanation:

the change in position is from 45cm to 27 cm (moving towards the negative x direction)

\Delta x = 45 cm - 27 cm = 18 cm

And the change in time:

\Delta t= 3 s

Now we must define the difference between speed and velocity:

Speed is a scalar quantity, which means that it is a number. Velocity ​​is also a number but you must also indicate the direction of the movement.

Thus, the speed is:

speed= \Delta x/ \Delta t = 18cm/3s=6cm/s

An the velocity is:

6cm/s in the negative direction

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Explanation:

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3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
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Answer:

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V=5m/sec

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The work energy principle states that the change in kinetic energy of an object is equal to the net work done on the object. If
sattari [20]

Answer:

v=\sqrt{2gh}\ m/s

Explanation:

From work energy theorem

Work done by all forces = Change in kinetic energy

Lets take

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h=height from the ground surface

initial velocity of object = 0 m/s

The final velocity of object is v

Work done by gravitational force = m g . h

The final kinetic energy = 1/2 m v²

So

Work done by all forces = Change in kinetic energy

m g h =  1/2 m v² - 0

v² = 2 g h

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