Physics: The measurement of the distance the particles move from the resting point of the wave

The measurement of the distance the particles, №13159511, 13.02.2022 10:01

Physics

In an atom smasher, two particles collide head on at relativistic speeds. if the velocity of the first particle is 0.741c to the left, and the velocity of the second particle is 0.543c to the right (both as measured in the lab rest frame), how fast are the particles moving with respect to each other?

Step-by-step answer

P Answered by Master

$W_x = 0.9156\ c$

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

$W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}$

u_x = 0.543 c

v_x = - 0.741 c

$W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}$

$W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}$

$W_x = \dfrac{1.284c}{1+0.402363}$

$W_x = 0.9156\ c$

Relative velocity of the particle is $W_x = 0.9156\ c$

Physics

In an atom smasher, two particles collide head on at relativistic speeds. if the velocity of the first particle is 0.741c to the left, and the velocity of the second particle is 0.543c to the right (both as measured in the lab rest frame), how fast are the particles moving with respect to each other?

Step-by-step answer

P Answered by Specialist

$W_x = 0.9156\ c$

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

$W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}$

u_x = 0.543 c

v_x = - 0.741 c

$W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}$

$W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}$

$W_x = \dfrac{1.284c}{1+0.402363}$

$W_x = 0.9156\ c$

Relative velocity of the particle is $W_x = 0.9156\ c$

Physics

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. (If an answer does not exist, enter DNE.) f(t) = t^3 − 8t^2 + 28t a. Find the velocity at time t. b. What is the velocity after 1 second? c. When is the particle at rest? d. When is the particle moving in the positive direction? e. Find the total distance traveled during the first 6 seconds.

Step-by-step answer

P Answered by PhD

6

(a) The velocity at time t, is 3t² - 16t + 28.

(b) The velocity after 1 second is 15 ft/s.

(c) The time when the particle is at rest does not exist.

(d) the particle is always moving in positive direction.

(e) The total distance traveled during first 6 seconds is 96 feet.

The given equation of motion;

f(t) = t³ - 8t² + 28t

(a) The velocity at time t, is calculated as follows;

$v = \frac{df}{dt} = 3t^2 -16t + 28$

(b) The velocity after 1 second is calculated as follows;

v = 3(1)² - 16(1) + 28

v = 15 ft/s

(c) When the particles is at rest, the velocity is zero;

3t² -16t + 28 = 0

solve the quadratic equation using formula method;

a = 3, b = -16, c = 28

$t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(-16) \ \ +/- \ \ \sqrt{(-16)^2 - 4(3\times 28)} }{2(3)} \\\\(-16)^2 \$

Thus, the time when the particle is at rest does not exist.

(d) The time of motion of the is greater than or equal to zero.

t ≥ 0

when t = 0

f(0) = (0)³ - 8(0)² + 28(0)

f(0) = 0 - 0 + 0

f(0) = 0

when t = 1

f(1) = (1)³ - 8(1) + 28(1)

f(1) = 1 - 8 + 28

f(1) = 21 feet

Thus, the particle started from zero origin and moves towards positive direction.

(e) The total distance traveled during first 6 seconds is calculated as follows;

f(6) = (6)³ - 8(6)² + 28(6)

f(6) = 96 feet.

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Physics

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = t3 − 12t2 + 36t (a) Find the velocity at time t. v(t) = 3t2−24t+36 (b) What is the velocity after 5 s? v(5) = ft/s (c) When is the particle at rest? (Enter your answer as a comma-separated list.) t = 6,2 (d) When is the particle moving in the positive direction? (Enter your answer in interval notation.) t = (2,6) (e) Find the total distance traveled during the first 8 s. ft

Step-by-step answer

P Answered by Master

a) v=3t^2-24t+36 is the velocity of the particle at any time t.

b) $v_5=87\ ft.s^{-1}$

c) The particle is at rest at time t=2 seconds.

d) At time t=2 seconds the particle posses positive velocity being at positive direction.

e) $s_8=32\ ft$

Explanation:

Given:

The function of displacement dependent on time, s=t^3-12t^2+36t .......(1)

a)

Now as we know that velocity is the time derivative of the displacement:

$v=\frac{d}{dt} s$

$v=\frac{d}{dt} (t^3-12t^2+36t)$

v=3t^2-24t+36 ..........................(2)

b)

Now the velocity after 5 seconds:

put t=5 in eq. (2)

$v_5=3\times 5^2-24\times 5+36$

$v_5=87\ ft.s^{-1}$

c)

When the particle is at rest it has zero velocity.

Now velocity at time t=6 s:

$v_6=3\times 6^2-24\times 6+36$

$v_6=120\ ft.s^{-1}$

Now velocity at time t=2 s:

$v_2=3\times 2^2-24\times 2+36$

$v_2=0\ ft.s^{-1}$

The particle is at rest at time t=2 seconds.

d)

Put the value t=2 sec. in eq. (1):

$s_2=2^3-12\times 2^2+36\times 2$

$s_2=32\ ft$

At time t=2 seconds the particle posses positive velocity being at positive direction.

e)

Distance travelled during the first 8 seconds:

put t=8 in eq. (1)

$s_8=8^3-12\times 8^2+36\times 8$

$s_8=32\ ft$

Physics

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. (If an answer does not exist, enter DNE.) f(t) = t^3 − 8t^2 + 28t a. Find the velocity at time t. b. What is the velocity after 1 second? c. When is the particle at rest? d. When is the particle moving in the positive direction? e. Find the total distance traveled during the first 6 seconds.

Step-by-step answer

P Answered by PhD

6

(a) The velocity at time t, is 3t² - 16t + 28.

(b) The velocity after 1 second is 15 ft/s.

(c) The time when the particle is at rest does not exist.

(d) the particle is always moving in positive direction.

(e) The total distance traveled during first 6 seconds is 96 feet.

The given equation of motion;

f(t) = t³ - 8t² + 28t

(a) The velocity at time t, is calculated as follows;

$v = \frac{df}{dt} = 3t^2 -16t + 28$

(b) The velocity after 1 second is calculated as follows;

v = 3(1)² - 16(1) + 28

v = 15 ft/s

(c) When the particles is at rest, the velocity is zero;

3t² -16t + 28 = 0

solve the quadratic equation using formula method;

a = 3, b = -16, c = 28

$t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(-16) \ \ +/- \ \ \sqrt{(-16)^2 - 4(3\times 28)} }{2(3)} \\\\(-16)^2 \$

Thus, the time when the particle is at rest does not exist.

(d) The time of motion of the is greater than or equal to zero.

t ≥ 0

when t = 0

f(0) = (0)³ - 8(0)² + 28(0)

f(0) = 0 - 0 + 0

f(0) = 0

when t = 1

f(1) = (1)³ - 8(1) + 28(1)

f(1) = 1 - 8 + 28

f(1) = 21 feet

Thus, the particle started from zero origin and moves towards positive direction.

(e) The total distance traveled during first 6 seconds is calculated as follows;

f(6) = (6)³ - 8(6)² + 28(6)

f(6) = 96 feet.

Learn more here:link

Physics

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = t3 − 12t2 + 36t (a) Find the velocity at time t. v(t) = 3t2−24t+36 (b) What is the velocity after 5 s? v(5) = ft/s (c) When is the particle at rest? (Enter your answer as a comma-separated list.) t = 6,2 (d) When is the particle moving in the positive direction? (Enter your answer in interval notation.) t = (2,6) (e) Find the total distance traveled during the first 8 s. ft

Step-by-step answer

P Answered by Master

a) v=3t^2-24t+36 is the velocity of the particle at any time t.

b) $v_5=87\ ft.s^{-1}$

c) The particle is at rest at time t=2 seconds.

d) At time t=2 seconds the particle posses positive velocity being at positive direction.

e) $s_8=32\ ft$

Explanation:

Given:

The function of displacement dependent on time, s=t^3-12t^2+36t .......(1)

a)

Now as we know that velocity is the time derivative of the displacement:

$v=\frac{d}{dt} s$

$v=\frac{d}{dt} (t^3-12t^2+36t)$

v=3t^2-24t+36 ..........................(2)

b)

Now the velocity after 5 seconds:

put t=5 in eq. (2)

$v_5=3\times 5^2-24\times 5+36$

$v_5=87\ ft.s^{-1}$

c)

When the particle is at rest it has zero velocity.

Now velocity at time t=6 s:

$v_6=3\times 6^2-24\times 6+36$

$v_6=120\ ft.s^{-1}$

Now velocity at time t=2 s:

$v_2=3\times 2^2-24\times 2+36$

$v_2=0\ ft.s^{-1}$

The particle is at rest at time t=2 seconds.

d)

Put the value t=2 sec. in eq. (1):

$s_2=2^3-12\times 2^2+36\times 2$

$s_2=32\ ft$

At time t=2 seconds the particle posses positive velocity being at positive direction.

e)

Distance travelled during the first 8 seconds:

put t=8 in eq. (1)

$s_8=8^3-12\times 8^2+36\times 8$

$s_8=32\ ft$

Physics

Aparticle moves according to the law of motion s(t) = t^{3}-8t^{2}+2t, t ge 0, where t is measured in seconds and s in feet. a.) find the velocity at time t. b.) what is the velocity after 3 seconds?c.) when is the particle at rest? enter your answer as a comma separated list. enter none if the particle is never at rest. at t_1= and t_2= with t_1

Step-by-step answer

P Answered by PhD

Explanation:

Given

displacement is given by

s(t)=t^3-8t^2+2t

so velocity is given by

$v(t)=\frac{\mathrm{d} s(t)}{\mathrm{d} t}$

v(t)=3t^2-16t+2

(b)velocity after t=3 s

$v(3)=3(3)^2-8\cdot 3+2$

v(3)=19 m/s

(c)Particle is at rest

when its velocity will become zero

v(t)=0

i.e. 3t^2-16t+2=0

$t=\frac{16\pm \sqrt{16^2-4\cdot 3\cdot 2}}{2\cdot 3}$

$t=\frac{16\pm 15.23}{6}$

t=5.20 s

Physics

Aparticle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.02t3 (a) find the velocity at time t (in ft/s). v(t) = .04t3−.06t2 (b) what is the velocity after 1 second(s)? v(1) = -.02 ft/s (c) when is the particle at rest? t = s (smaller value) t = s (larger value) (d) when is the particle moving in the positive direction? (enter your answer using interval notation.) (e) find the total distance traveled during the first 12 seconds. (round your answer to two decimal places.) ft (f)

Step-by-step answer

P Answered by PhD

1

Since you solved a and b I will start with part c.
Part C
To answer this question we need to find zeros of a velocity function:
v(t)=0.04t^3-0.06t^2

We can factor this polynomial:
v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)

Now it's pretty easy to find zeros. This function will be equal to zero when any of the factors are equal zero.
$t^2=0;\\ 0.04t-0.06=0$
We solve these two equations and we get our zeros:
$t_1=0; t_2=\frac{3}{2}$
The particle is at rest at t=0 and t=3/2.
Part D
To solve this we need to determine when our velocity function is greater than zero. We will use factored form.
We determine when each factor is greater than zero and with that information, we build the following table:
$\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}$
We can see, from the table, that our function is positive when $- \infty < t$ and t>3/2.
That is the range in which particle is moving in positive direction.
Part E
We know that distance traveled is given with:
s(t)=0.01t^4 - 0.02t^3

We simply plug in t=12 to find total distance traveled:
s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft

Part F
We know that acceleration is defined as a rate of change of velocity.
We find acceleration by taking the first derivative of velocity with respect to time.
$a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t$
To find acceleration after 1 second we simply plug in t=1s in above equation:
a(1)=0.12-0.12=0

Physics

Aparticle moves according to the law of motion s(t) = t^{3}-8t^{2}+2t, t ge 0, where t is measured in seconds and s in feet. a.) find the velocity at time t. b.) what is the velocity after 3 seconds?c.) when is the particle at rest? enter your answer as a comma separated list. enter none if the particle is never at rest. at t_1= and t_2= with t_1

Step-by-step answer

P Answered by PhD

Explanation:

Given

displacement is given by

s(t)=t^3-8t^2+2t

so velocity is given by

$v(t)=\frac{\mathrm{d} s(t)}{\mathrm{d} t}$

v(t)=3t^2-16t+2

(b)velocity after t=3 s

$v(3)=3(3)^2-8\cdot 3+2$

v(3)=19 m/s

(c)Particle is at rest

when its velocity will become zero

v(t)=0

i.e. 3t^2-16t+2=0

$t=\frac{16\pm \sqrt{16^2-4\cdot 3\cdot 2}}{2\cdot 3}$

$t=\frac{16\pm 15.23}{6}$

t=5.20 s

Physics

Aparticle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.02t3 (a) find the velocity at time t (in ft/s). v(t) = .04t3−.06t2 (b) what is the velocity after 1 second(s)? v(1) = -.02 ft/s (c) when is the particle at rest? t = s (smaller value) t = s (larger value) (d) when is the particle moving in the positive direction? (enter your answer using interval notation.) (e) find the total distance traveled during the first 12 seconds. (round your answer to two decimal places.) ft (f)

Step-by-step answer

P Answered by PhD

1

Since you solved a and b I will start with part c.
Part C
To answer this question we need to find zeros of a velocity function:
v(t)=0.04t^3-0.06t^2

We can factor this polynomial:
v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)

Now it's pretty easy to find zeros. This function will be equal to zero when any of the factors are equal zero.
$t^2=0;\\ 0.04t-0.06=0$
We solve these two equations and we get our zeros:
$t_1=0; t_2=\frac{3}{2}$
The particle is at rest at t=0 and t=3/2.
Part D
To solve this we need to determine when our velocity function is greater than zero. We will use factored form.
We determine when each factor is greater than zero and with that information, we build the following table:
$\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}$
We can see, from the table, that our function is positive when $- \infty < t$ and t>3/2.
That is the range in which particle is moving in positive direction.
Part E
We know that distance traveled is given with:
s(t)=0.01t^4 - 0.02t^3

We simply plug in t=12 to find total distance traveled:
s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft

Part F
We know that acceleration is defined as a rate of change of velocity.
We find acceleration by taking the first derivative of velocity with respect to time.
$a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t$
To find acceleration after 1 second we simply plug in t=1s in above equation:
a(1)=0.12-0.12=0

The measurement of the distance the particles move from the resting point of the wave

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