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A particle's location(x)  is in a relationship with the time(t).The relation is x = at^2-bt^3.What will the paticle's acceleration after  2 second?

 Apr 24, 2016

Best Answer 

 #2
avatar+33661 
+10

I should have used a different letter for acceleration above, so as not to confuse it with the constant  a.

 

Acceleration better as, say:    acc = 2a - 6bt.

.

 Apr 24, 2016
 #1
avatar+33661 
+5

Displacement:   x = at^2 - bt^3

 

Velocity:            v = 2at - 3bt^2                  Velocity is rate of change of displacement with time

 

Acceleration:    a = 2a - 6bt                       Acceleration is rate of change of velocity with time.

.

 Apr 24, 2016
 #2
avatar+33661 
+10
Best Answer

I should have used a different letter for acceleration above, so as not to confuse it with the constant  a.

 

Acceleration better as, say:    acc = 2a - 6bt.

.

Alan Apr 24, 2016
 #4
avatar+257 
0

Thank You ALAN!.

In our Indian style Acceleration better as, say: a. I don't want to say that you are wrong.smileylaugh

AaratrikRoy  Apr 24, 2016
 #3
avatar+37146 
+9

 x = at^2-bt^3  

The first derivative will be the velocity  v=  2at - 3bt^2 

The second derivative will be the acceleration= 2a - 6bt    

   SInce t= 2 in this question,  Acceleration(@t=2) = 2a-12b

 Apr 24, 2016
 #5
avatar+257 
0

Guest, you are right. But your last line isn't truthful. 

AaratrikRoy  Apr 24, 2016
 #6
avatar+257 
0

You are right .Thank You ElectricPavlov

AaratrikRoy  Apr 27, 2016

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