The position of a particle is given by the function

Asked by maham237 @ in Physics viewed by 299 People

x=(4t3−6t2+12)m, where t is in s. A.) at what time does the particle reach its minimum velocity
B.) what is (vx)min
C.) at what time is the acceleration zero

Answered by maham237 @


Answer

given,

x = 4 t³ - 6 t² + 12

velocity, v = \dfrac{dx}{dt}

\dfrac{dx}{dt}=\dfrac{d}{dt}(4t^3-6t^2+12)

v =12t^2-12t

For minimum velocity calculation we have differentiate it and put it equal to zero.

\dfrac{dv}{dt} =\dfrac{d}{dt}12t^2-12t

\dfrac{dv}{dt} =24t-12.........(1)

putting it equal to zero

24 t - 12 =0

t = 0.5 s

At t = 0.5 s velocity will be minimum.

b) minimum velocity

v = 12t² -12 t

v = 12 x 0.5² -12 x 0.5

v = -3 m/s

c) derivative of velocity w.r.t. time is acceleration

from equation 1

a = 24 t - 12

time at which acceleration will be zero

0 = 24 t - 12

t = 0.5 s

At t = 0.5 s acceleration will be zero.


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