Question

The velocity of a particle traveling in a straight line is given by v = (6t - 3t²) m/s, where t is in seconds. If s = 0 when t = 0, determine the particle’s deceleration and position when t = 3s. How far has the particle traveled during the 3-s time interval, and what is its average speed?

Answer 1

Answer

given,

v = (6 t - 3 t²) m/s

we know,

$v = \dfrac{dx}{dt}$

$a = \dfrac{dv}{dt}$

position of the particle

$dx = v dt$

integrating both side

$\int dx = (6t - 3 t^2)\int dt$

 x = 3 t² - t³

Position of the particle at t= 3 s

 x = 3  x 3² - 3³

 x = 0 m

now, particle’s deceleration

$a = \dfrac{dv}{dt}$

$a = \dfrac{d}{dt}(6t - 3 t^2)$

  a = 6 - 6 t

at t= 3 s

    a = 6 - 6 x 3

    a = -12 m/s²

distance traveled by the particle

  x = 3 t² - t³

at t = 0 x = 0

   t = 1 s   , x = 3 (1)² - 1³ = 2 m

   t = 2 s  ,  x = 3(2)² - 2³ = 4 m

   t = 3 s ,   x =  0 m

total distance traveled by the particle

D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s

D = 2 + 4 + 2 = 8 m

average speed of the particle

$v_{avg} = \dfrac{distance}{time}$

$v_{avg} = \dfrac{8}{3}$

$v_{avg} =2.67\ m/s$