The position of a particle is given by the function 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

Answers

The reflection of the candle in the concave mirror formed an upside-down virtual image that is smaller than the candle.in other words. the answer is the third option, c

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.

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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