Question

❎CHALLANGE❎
Motion of a particle along a straight line is described by equation X= 8 + 12t - t^3
Where X is in metre and t is in second, What is retardation of particle when its velocity becomes zero?
Please give full solution...

Answer 1

Heya........!!!!

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- Given Equation :-- x = 8 + 12t - t^3

➡ Differentiating the equation , dx/dt .

➡ dx/dt = velocity = 0 + 12 - 3t^2 .

➡ Velocity (v ) = 12 - 3t^2

➡ Velocity becomes 0 so equate the equation of velocity = 0

➡ 12 - 3t^2 = 0

➡ 3t^2 = 12

➡ t^2 = 4

➡ t = ± 2 ,, ( + 2 , - 2 )

➡ Differentiating the equation of Velocity we get acceleration

➡ dv/dt = a. = 0 - 6t

➡ putting the value of t we get Retardation

➡ a = - 6t

➡ a = - 6 × 2

➡ a = - 12 m/s^2 ( retardation ) .


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Hope It Helps You ^_^

Answer 2

Heya!

X = 8 + 12t - t³
v = dx/dt = (0) + 12 - 3t²

When, v = 0

12 - 3t² = 0
12 = 3t²
t = 2 sec.

Now, a = dv/dt = -6t

So, retardation = 12 m/s²