Get Help on Integral Calculus: Velocity & Distance Equation

[franz32]

I need guidance (updated)

The website has changed a lot! Very beautiful and attactive! I like it.

Well, I need help. This is on integral calculus . How do I find an equation involving the velocity "v" and the distance "s" given that the acceleration "a" is 800; and that v = 20 when s = 1?

 

[philosophking]

Velocity(v) is the integral of the acceleration function. Position(s) is the integral of the velocity function. The variable that you're integrating with respect to is t. I hope this helps you, it's pretty easy from there =P

 

[Inspector Gadget]

s(t) = position
s'(t) = v(t) = velocity
s''(t) = v'(t) = a(t) = acceleration

(EDIT: Sorry...See below...I made a mistake).

 

[philosophking]

Yes, you have to say v(t) = 20 and s(t) = 1

 

[philosophking]

After trying this problem for quite some time, I have realized that in its present wording, it is impossible to solve :)

 

[HallsofIvy]

After trying this problem for quite some time, I have realized that in its present wording, it is impossible to solve :)

Right. There will be two unknowns and only one additional condition.

Since the acceleration is a constant 800, the velocity at any time t is
v(t)= 800t+ v₀ where v₀ is the (unkown) velocity at time t=0. The position is s(t)= 400t²+ v₀t+ s₀ where s₀ is the (unknown) position at time t= 0.

Knowing that v(t)= 20 and s(t)= 1 for some t allows us to reduce to only one unknown but not get rid of both. If we knew what that "t" was, then we could answer this question.

 

[Nexus[Free-DC]]

Wouldn't the sensible thing be to add a disclaimer like, "...assuming the starting point to be zero, we find that..."

 

[franz32]

Thank you for all of ur helps. =)

 

[franz32]

Oh... I think there's another way one can solve it:

a = dv/dt = dv/ds X ds/dt = v X dv/ds