Aparticle's position along the x-axis is described by. x(t)= at+bt^2where t is in seconds: x is in meters: and the constants a and b are given below. randomized variables a= -3.5 m/s b= 3.9 m/s^2 a. what is the velocity, in meters per second. of the particle at the time t1= 3.0 s? b. what is the velocity, in meters per second: of the particle when it is at the origm (x=0) at time to> 0?

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.

------(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,

   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

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2