Prove that acceleration is constant

Hello,

Finding my self very frustrated while working on my calculus homework... seeing if someone can help me through where i am getting stuck!

Question :

An object moves so that its velocity, v, is related to its position, s, according to the function v=sqrt(b^2+2gs) where b and g are constants. Show that the acceleration (with respect to time) of the object is constant.

Thank you!, i would really appreciate it, its the only question im stuck on. (Headbang)

Re: Prove that acceleration is constant

Quote:

Originally Posted by

**elittlewood** Hello,

Finding my self very frustrated while working on my calculus homework... seeing if someone can help me through where i am getting stuck!

Question :

An object moves so that its velocity, v, is related to its position, s, according to the function v=sqrt(b^2+2gs) where b and g are constants. Show that the acceleration (with respect to time) of the object is constant.

Thank you!, i would really appreciate it, its the only question im stuck on. (Headbang)

There are two ways. The first is to take the time derivative of v (which is a. Note that ds/dt = v.) Then you have the equation:

$\displaystyle \frac{dv}{dt} = \frac{1}{2} \cdot \frac{1}{\sqrt{b^2 + 2gs}} \cdot 2gv$

How can you simplify this? (Hint: Look at the square root. What is this equal to?)

Or you could look at the Physics equation:

$\displaystyle v^2 = v_0^2 + 2as$

where v_0 = b and g = a...

-Dan

Re: Prove that acceleration is constant

Hello thanks for the quick reply !

Okay, i see what you are doing, but let me clarify so i don't get confused. You have used the chain rule to differentiate v. Now, when you say simply this is where i get a little confused, you have pointed out the sqrt... it is equal to velocity so can you sub the v in ? Would that be the next step?

Re: Prove that acceleration is constant

Quote:

Originally Posted by

**elittlewood** Hello thanks for the quick reply !

Okay, i see what you are doing, but let me clarify so i don't get confused. You have used the chain rule to differentiate v. Now, when you say simply this is where i get a little confused, you have pointed out the sqrt... it is equal to velocity so can you sub the v in ? Would that be the next step?

You got it! :) Then do some simple simplifications.

-Dan

Re: Prove that acceleration is constant

Thank you! i simplified and was left with g...

May i ask when you first differentiated you had at the end 2gv . Where did the v come from ?

Re: Prove that acceleration is constant

Quote:

Originally Posted by

**elittlewood** Thank you! i simplified and was left with g...

May i ask when you first differentiated you had at the end 2gv . Where did the v come from ?

Sorry.

$\displaystyle \frac{dv}{dt} = \frac{1}{2} \cdot \frac{1}{\sqrt{b^2 + 2gs}} \cdot \frac{d}{dt}(b^2 + 2gs) = \frac{1}{2} \cdot \frac{1}{\sqrt{b^2 + 2gx}} \cdot \frac{d}{dt}(2gs)$

And of course ds/dt = v.

-Dan

Re: Prove that acceleration is constant

V = √(b^2+2gs) = ds/dt [ Velocity ]

(d^2 s)/(dt^2 )= 1/(2 √(b^2+2gs) ) 2g (ds )/dt = 1/(2 √(b^2+2gs) ) 2g √(b^2+2gs) = g that is constant and hence the acceleration is constant