Respuesta :
Answer
given,
position of particle
x(t)= A t + B t²
A = -3.5 m/s
B = 3.9 m/s²
t = 3 s
a) x(t)= -3.5 t + 3.9 t²
velocity of the particle is equal to the differentiation of position w.r.t. time.
[tex]\dfrac{dx}{dt}=\dfrac{d}{dt}(-3.5t + 3.9t^2)[/tex]
[tex]v= -3.5 + 7.8 t [/tex]------(1)
velocity of the particle at t = 3 s
v = -3.5 + 7.8 x 3
v = 19.9 m/s
b) velocity of the particle at origin
time at which particle is at origin
x(t)= -3.5 t + 3.9 t²
0 = t (-3.5 + 3.9 t )
t = 0, [tex]t=\dfrac{3.5}{3.9}[/tex]
t = 0 , 0.897 s
speed of the particle at t = 0.897 s
from equation (1)
v = -3.9 + 7.8 t
v = -3.9 + 7.8 x 0.897
v = 3.1 m/s
To solve the problem we should know about velocity.
Velocity
Velocity is the rate of change of its position with respect to time.
[tex]V = \dfrac{dy}{dt}[/tex]
Given to us
- x(t)= At+Bt^2
- A= -3.5 m/s
- B= 3.9 m/s^2
Velocity of Particle
x(t)= At+Bt²
[tex]V(t) = \dfrac{dy}{dt} = \dfrac{d(At+Bt^2)}{dt} = A+2Bt[/tex]
A.) the velocity, in meters per second. of the particle at the time t1= 3.0 s,
Velocity of particle(t = 3.0 s)
[tex]V(t) = A +2Bt[/tex]
Substituting the values,
[tex]V(t_1=3) = (-3.5) +2(3.9)(3.0)\\\\V(t_1=3) = 19.9\ m/s[/tex]
B.) the velocity, in meters per second: of the particle when it is at the origin (x=0) at t ≥ 0
Displacement, x = 0
[tex]x(t)= At+Bt^2\\\\0 = At+Bt^2\\\\[/tex]
Taking t as common,
[tex]0 = t(A+Bt)\\\\[/tex]
[tex]0 = (A+Bt)\\\\[/tex]
Substituting the values and solving or t,
[tex]0 = A+ Bt\\0 = -3.5 + (3.9)t\\3.5=3.9t\\t= \dfrac{3.5}{3.9}\\\\t = 0.8974\ s[/tex]
Velocity of particle(t = 0.8974 s)
Substituting the value in the formula of velocity,
[tex]V(t) = A +2Bt[/tex]
Substituting the values,
[tex]V(t_1=0.8974) = (-3.5) +2(3.9)(0.8974)\\\\V(t_1=3) = 3.5 \ m/s[/tex]
Hence, the velocity of the particle is 3.5 m/s.
Learn more about Velocity:
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