Solving for variables

• marissa12
In summary, the particle accelerates from v_initial to vfinal, undergoing displacement s given by s=x_final-x_initial. During a certain interval of time, the particle accelerates from v_initial to vfinal, undergoing displacement s given by s=x_final-x_initial. The acceleration is in terms of v_final, v_initial, and s. The work in terms of m, v_initial, and v_final is W=F*s. The s is still in the equation though. The 2nd part of the problem is w=intergral of { mv*dv} between v_final and v_initial. and we are supposed to express that inf

marissa12

a particle of mass m moving in the x direction at constant acceleration a . During a certain interval of time, the particle accelerates from v_initial to vfinal , undergoing displacement s given by s=x_final-x_initial.

Express the acceleration in terms of v_initial, v_final, and s:
that was easy.. i got

a=(v_final-v_initial)/(s/((v_initial+v_final)/2))

since F=ma
W=F*s

the question is Give an expression for the work in terms of m, v_initial, v_final ?

and i can't seem to eliminate the 's' because i end up getting

W=m*a(see above) *s and i can't get eliminate it? any ideas?

the 2nd part of the problem is w= intergral of { mv*dv} between v_final and v_initial
and we are supposed to express that in m_vinitial and v_final

and i got the answer ((m*v_final)^2-(m*v_initial)^2)/2 and that didnt work. i used the basic intergral formula

fint{t*dt between a and b} = b^2-a^2/t

marissa12 said:
...
the question is Give an expression for the work in terms of m, v_initial, v_final ?

and i can't seem to eliminate the 's' because i end up getting

W=m*a(see above) *s and i can't get eliminate it? any ideas?
can't you use ,

$$v_f^2 - v_i^2 = 2as$$ ?

the s is still in the equation though

marissa12 said:
the 2nd part of the problem is w= intergral of { mv*dv} between v_final and v_initial
and we are supposed to express that in m_vinitial and v_final

and i got the answer ((m*v_final)^2-(m*v_initial)^2)/2 and that didnt work. i used the basic intergral formula

fint{t*dt between a and b} = b^2-a^2/t
$$\int_{vi}^{vf} mv dv = m[v^2/2]_{vi}^{vf}$$

The m doesn't get squared.

marissa12 said:
the s is still in the equation though
substitute for "as" from one eqn into t'other.

oo that's right stupid parentheses.. but i still can't get the first part?

W = Fs = Mas

vf² - vi² = 2as ==> as = ½(vf² - vi²)

So,

W = Mas = M*½(vf² - vi²)
W = (M/2)(vf² - vi²)
================

Last edited:
o wow yea that makes a lot more sense than what i had, i used the wrong eq. lol thank youuu!