Question

A particle'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 1

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.

$\dfrac{dx}{dt}=\dfrac{d}{dt}(-3.5t + 3.9t^2)$

$v= -3.5 + 7.8 t$------(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, $t=\dfrac{3.5}{3.9}$

   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