Question

Vf = V0 + at

Xf= x0 + V0t + 1/2at^2

V^2f = V^20 +2a (xf - X0)

1. Using equation (1) solve for t given that v0 = 5 m/s, vf = 25 m/s, and a = 10 m/s2.

2. Given v0 = 0 m/s, x0 = 0 m and t = 10 s, use all three equations together to find xf .

3. a = 10 m/s2, x0 = 0 m, xf = 120 m, and v0 = 20 m/s. Use the second equation to find t.

4. vf = - v0 and a = 2 m/s2. Use the first equation to find t / 2.

5. How does each equation simplify when a = 0 m/s2 and x0 = 0 m?

Answer 1

Step-by-step explanation:

1., 3., 4., and 5. all tell you exactly what to do and give you the values to do it, so I'm not sure what your issue for those. It's plug and chug for those four.

As for 2., something is missing here... I've tried several different ways of combining the three equations (which is what you would do there) but can't get any values out of it without knowing either V_f or acceleration. Double check your original question to see if you might have missed a given variable and update us on it.

Edit: Or maybe they want you to answer with variables? I don't think that's the case though; because why give a value for time if that's what they're looking for...

Answer 2

1. t= 2s

2.
`x_(f) =(1)/(2) * a * t^(2)`

3. 3.29 s

4.
`(t)/(2) = (V_(f) )/(a)`

5. a)
`V_(f) = V_(o)`

b)
`V_(f) =V_(o)`

c)
`V_(f) =V_(o)`

Step-by-step explanation:

1.

`V_(o) = 5 (m)/(s)`,
`V_(f) = 25 (m)/(s)`,
`a= 10 (m)/(s^(2) )`

`V_(f) = V_(o) + a*t\\a*t=V_(f) - V_(o) \\t= (V_(f) - V_(o))/(a)\\ t=(25-5 (m)/(s) )/(10 (m)/(s^(2) ) ) = (20(m)/(s) )/(10 (m)/(s^(2) ) )\\t= 2`s

2.

a.
`V_(f) = V_(o) + a*t`

`V_(f) = a*t`

c.
`V_(f) ^(2) =V_(o) ^(2) +2*a* (x_(f) -x_(o))`

`V_(f) ^(2) = 2*a*x_(f)`

`(a*t) ^(2) = 2*a*x_(f)`

`a ^(2)*t ^(2) = 2*a*x_(f)`

`a *t ^(2) = 2*x_(f)`

`x_(f) =(1)/(2) * a * t^(2)`

3.

`a= 10 (m)/(s^(2) )`,
`x_(o) = 0m`,
`x_(f) = 120m`,
`V_(o) = 20 (m)/(s)`

`x_(f) = x_(o)+V_(o)*t + (1)/(2) * a * t^(2) \\120m =0m + 20(m)/(s)*t + (1)/(2)*10(m)/(s^(2) )  *t^(2) \\5t^(2) +  20(m)/(s)*t -120=0\\t^(2) +  4(m)/(s)*t -24=0\\t_(1,2) =  \frac{-b +/- \sqrt{b^(2)-4*a*c}}{2*a} \\t_(1) =  \frac{-4 + \sqrt{4^(2)-4*1*-24}}{2*1}= 3.29 \\t_(2) =  \frac{-4 - \sqrt{4^(2)-4*1*-24}}{2*1}= -7.29`

No negative time so t1 is the useful, t=3.29s

4.

`V_(f) = -V_(o)`,
`a= 2 (m)/(s^(2) )`

`V_(f) = V_(o) + a*t`

`V_(f) = -V_(f) +a*t\\ V_(f)+V_(f)= a * t\\t=(V_(f)+V_(f))/(a) =(2*V_(f) )/(a)\\(t)/(2)=(V_(f) )/(a)`

5.

`x_(o) = 0m`,
`a= 0 (m)/(s^(2) )`

a).

`V_(f) = V_(o) + a*t \\ V_(f) = V_(o) + 0*t \\V_(f) = V_(o)`

b).

`x_(f) =x_(o)+ V_(o)*t + (1)/(2) * a * t^(2) \\x_(f) =0 + V_(o)*t + (1)/(2) * 0 * t^(2) \\ \\x_(f) =V_(o)*t \\\\(x_(f) )/(t) =V_(o) \\V_(f) =V_(o)`

c).

`V_(f) ^(2) =V_(o) ^(2) +2*a* (x_(f) -x_(o))\\ V_(f) ^(2) =V_(o) ^(2) +2*0* (x_(f) -x_(o))\\ V_(f) ^(2) =V_(o) ^(2) \\\sqrt{V_(f) ^(2)} =\sqrt{V_(o) ^(2)}\\ V_(f)  =V_(o)`

The motion without any acceleration is the same speed according to law of Newton