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3 years ago

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A particle moves along a straight line such that its position is defined by s = (t^3- 3t^2 2) m. Determine the average velocity,

the average speed, and the acceleration of the particle when t = 4 s.

1 answer:

sergey [27]3 years ago

7 0

From the information in the question, the velocity of the particle is 24 ms-1.

Let us recall that the velocity of a body is obtained as the first derivative of distance covered with respect to time. Hence;

ds/dt = v

Now, we have to differentiate (t^3- 3t^2 + 2)m with respect to t and we have;

d(t^3- 3t^2 + 2)m/dt = 3t^2 - 6t

Now at t= 4s, the velocity of the particle is;

3(4)^2 - 6(4) = 24 m/s

The acceleration of the particle is the second derivative of distance with respect to time;

d^2s/dt^2 = d^2 3t^2 - 6t/dt^2 = 6t

Substituting t = 4s

a = 24 ms-2

Learn more: brainly.com/question/24486060

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Explanation:

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3 years ago

A child swings on a playground swing. How many times does the child swing through the swing's equilibrium position during the co

Karolina [17]

<h2>The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>

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An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.7 s. A passenger in the elevator i

avanturin [10]

Answer: Tension = 53.6N

Explanation:

Given that

Height h = 1 m

Time t = 1.7 s.

Mass m = 5.1 kg

From the equation of the motion we can get the acceleration of the elevator:

h = X0+ V0t + at2/2;

Th elevator starts from rest with a constant upward acceleration. Initial velocity Vo = 0, also Xo = 0; thus

a = 2h/t2 = 2 × 1/1.7^2

a = 0.69 m/s2.

Then we can find the tension in the cord by using the formula

T = mg + ma

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= 5.1 × 10.5

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3 years ago

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

$k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

$v_d$ is the velocity at distance d from the proton = 2 $v_i$

So the equation becomes

$\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

$\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

$3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

$d = \frac{2k(q)^2}{3mv_i^2}$

$= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

$=1.66*10^{-9}m$

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3 years ago