position position x of an object is related to time and x=t³-12t where x is in meter and t in seconds find the acceleration of the object when its velocity is 0???
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0
=18−6t
The time will be
0=18−6t
t=3
The position of the particle at 3 sec
Put the value of t in equation (I)
x=9×9−27
x=81−27
x=54 m
The position of the particle will be 54 m.
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4
Differentiate wrt time
dx/dt = 3t^2 - 12
dx/dt = v (velocity)
v = 3t^2 - 12 (1)
Time at which v = 0
3t^2 - 12 = 0
On solving t = 2 at which v = 0
Further differentiating equation (1)
dv/dt = 6t
dv/dt = a (acceleration)
a = 6t
Putting t = 2
a = 12m/s^2
dx/dt = 3t^2 - 12
dx/dt = v (velocity)
v = 3t^2 - 12 (1)
Time at which v = 0
3t^2 - 12 = 0
On solving t = 2 at which v = 0
Further differentiating equation (1)
dv/dt = 6t
dv/dt = a (acceleration)
a = 6t
Putting t = 2
a = 12m/s^2
anagha71:
can u plz explain how u get dv÷dt =6t
Differentiate equation 3t^2-12 wrt time
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