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A particle's position along the x-axis is described by: x(t)=A t + B t​2

where t is in seconds, x is in meters, and the constants A and B are given below.

​Randomized Variables

A = -4.3 m/s

B = 4.9 m/s​2

(a) What is the velocity, in meters per second, of the particle at t​1 = 3.0 s?
(b) What is the velocity, in meters per second, of the particle when it is at the origin (x=0) at t0 > 0?

Respuesta :

Answer:

a)V = 25.1 m/s

b)V = 4.226 m/s

Explanation:

Given that

x(t)=A t + B t​²

A = -4.3 m/s

B = 4.9 m/s​²

x(t)=  - 4.3 t +4.9 t​²

The velocity of the particle is given as

[tex]V=\dfrac{dx}{dt}[/tex]

V=-4.3 + 4.9 x 2 t

V= - 4.3 + 9.8  t m/s

Velocity at point t= 3 s

V= - 4. 3 + 9.8 x 3 m/s

V= - 4.3 + 29 .4 m/s

V = 25.1 m/s

At origin :

x= 0 m

0 =  - 4.3 t +4.9 t​²

0 = - 4.3 + 4.9 t

[tex]t=\dfrac{4.3}{4.9}\ s[/tex]

t=0.87 s

The velocity at t= 0.87 s

V= - 4.3 + 9.8  t m/s

V= - 4. 3 + 9.8 x 0.87 m/s

V= - 4.3 + 8.526 m/s

V = 4.226 m/s

a)V = 25.1 m/s

b)V = 4.226 m/s

onyebuchinnaji

The velocity of the particle at t = 3.0s is 25.1 m/s.

The velocity of the particle when it is at the origin is 4.324 m/s.

The given parameters:

  • Position of the particle, x(t) = At + Bt²
  • A = -4.3 m/s and B = 4.9 m/s²

The velocity of the particle at t = 3.0s is calculated as follows;

[tex]v = \frac{dx}{dt} = A + 2Bt\\\\v = -4.3 + 2(4.9\times 3)\\\\v = 25.1 \ m/s[/tex]

The velocity of the particle when it is at the origin (x = 0)

[tex]0 = -4.3t + 4.9t^2\\\\0 = t(-4.3 + 4.9t)\\\\t = 0 \ \ \ or \ \ -4.3 + 4.9t = 0\\\\4.9t = 4.3\\\\t = \frac{4.3}{4.9} \\\\t = 0.88 \ s\\\\v = A + 2Bt\\\\ v = -4.3 + (2\times 4.9 \times 0.88)\\\\v = 4.324 \ m/s[/tex]

Learn more about velocity of particle here: https://brainly.com/question/80295?source=archive

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