Help me anyone on physics?
3 Answers
We get
Explanation:
Given a point on the curve
We write this in terms of the components,
Integrating,
The launch point is the origin, so
Choice (1).
(1)
Explanation:
Given
As
#v_x=3#
We know that velocity means
#dx=3dt#
Integrating both sides with respective variables we get
#int\ dx=int\ 3dt#
#=>x=3t+c#
where#c# is constant of integration.
To evaluate
#x=3t# .........(1)
Similarly for
#v_y=6x#
We know that velocity means
#dy=6xdt#
Using (1)
#dy=6(3t)dt#
#dy=18tdt#
Integrating both sides with respective variables we get
#int\ dy=18int\ tdt#
#=>y=18xxt^2/2+c_1#
#=>y=9t^2+c_1#
where#c_1# is constant of integration.
To evaluate
#y=6t^2# .........(2)
Eliminating
#y=x^2#
Alternate solution in this case.
(1)
Explanation:
Given
As
#v_x=3#
Using definition of velocity we get
#dotx=3#
Integrating both sides with respect to time we get
#int\ dotx\ dt=int\ 3\ dt#
#=>x=3t+c#
where#c# is constant of integration.
To evaluate
#x=3t# .........(1)
Similarly for
#v_y=6x#
#=>dot v=6x#
Using (1)
#doty=6(3t)dt#
#=>doty=18t#
Integrating both sides with respect to time we get
#int\ doty\ dt=18int\ t\ dt#
#=>y=18xxt^2/2+c_1#
#=>y=9t^2+c_1#
where#c_1# is constant of integration.
To evaluate
#y=6t^2# .........(2)
Eliminating
#y=x^2#
.-.-.-.-.-.-.-.-.-
These solutions have been posted keeping in my that
#d/dt# in an expression#dx/dt# is an operator.
#because dx/dt# is actually#d/dt(x)#
Mathematicians feel that it is not good a idea to break the operator into numerator and denominator as we generally resort to in Physics problems.