the position x of a particle with respect to time t along x axis is givrn by x=9t^2 -t^3, where x is in meters and t in second what will be the position of this paticle when it achieves maximum speed along the positive x direction??
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Answered by
15
speed is derivative of position
hence f'(x)=18t-3t^2
f''(x)=18-6t
hence we have to find maxima
put 18t-3t^2=0
hence T=0 or T=6
for maxima 0
put in f''(x)
=18 ×3=54
hence f'(x)=18t-3t^2
f''(x)=18-6t
hence we have to find maxima
put 18t-3t^2=0
hence T=0 or T=6
for maxima 0
put in f''(x)
=18 ×3=54
Anonymous:
follow me 4 more answers
Answered by
24
Hey there!!
Position of particle is given by :
=> x = 9t²- t³ -----------------(1)
We have find max speed so we need to diffrentiate 'x' w.r.t time.
So, v = dx/dt = 18t - 3t² --------(2)
Now we have to find maximum speed so : dv/dt = 0
=> dv/dt = 18 - 6t = p
=> t = 18/6 = 3 sec.
Now speed will be maximum when t = 3 sec. So put t = 3 in 'x'
54
So position will be : X = 9(3)² - (3)³ = 54 m
_______________________________________
. \ option 3 is correct
Hope it will help u
Position of particle is given by :
=> x = 9t²- t³ -----------------(1)
We have find max speed so we need to diffrentiate 'x' w.r.t time.
So, v = dx/dt = 18t - 3t² --------(2)
Now we have to find maximum speed so : dv/dt = 0
=> dv/dt = 18 - 6t = p
=> t = 18/6 = 3 sec.
Now speed will be maximum when t = 3 sec. So put t = 3 in 'x'
54
So position will be : X = 9(3)² - (3)³ = 54 m
_______________________________________
. \ option 3 is correct
Hope it will help u
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