+257 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?
+33616 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.
.
+33616 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.
.
+257 Thank You ALAN!.
In our Indian style Acceleration better as, say: a. I don't want to say that you are wrong.
+36916 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
