Van travels over the hill described by y=(-1.5(1/1000)x^2 + 15)ft

[vanquish]

a=(75ft/s^2)ut+(v2/ρ)un
At=75ft/s^2 ?
An=v^2/ρ=75^2/344.6=16.32

 

[nvn]

a_n looks correct, but a_t is wrong. Look in your textbook for the definition of acceleration, and compute a_t.

 

[vanquish]

the only things my textbook says about this are:
a_t=\dot{v}
or
a_tds=vdv

 

[nvn]

OK, use the first equation to compute a_t.

 

[vanquish]

I think I got it, At=11.25
An=v^2/ρ=75^2/344.6=16.32

So when i calculated everything and used the angle i found before to project this information from the At and An directions into the Ax and Ay directions I get these numbers:
Ax=11.12
Ay=16.14

 

[nvn]

a_n looks correct, but a_t is wrong. Show your calculations for a_t.

 

[vanquish]

according to my book:
a_tds=vdv
so a_t=vdv/ds
v=75m/s and dv/ds=-3x/1000
and since we want At at 50ft
dv/ds=-.15
so at=(75)(-.15)

 

[nvn]

Statements by vanquish highlighted in blue.
a_t*ds = v_t*dv_t[/color] --true.
so a_t = v_t*dv_t/ds[/color] = (ds/dt)*dv_t/ds = dv_t/dt[/color] --true.
v_t = 75 m/s[/color] --true, except for units; also, see the note I added to post 30.
and dv/ds = -3x/1000[/color] --false; dy/dx = -0.003*x.
and since we want a_t at 50 ft[/color]
dv/ds = -0.15[/color] --false.
so a_t = 75(-0.15)[/color] --false.

By the way, according to the international standard (ISO 31-0), numbers less than 1 must have a zero before the decimal point. E.g., 0.15, not .15.

 

[nvn]

vanquish: Hint 6: If s = 75*t, can you compute ds/dt, and d²s/dt²?

 

[vanquish]

vanquish: Hint 6: If s = 75*t, can you compute ds/dt, and d²s/dt²?

ds/dt=75
d²s/dt²=0

 

[nvn]

That's correct. What is another name for ds/dt and d²s/dt²?

 

[vanquish]

That's correct. What is another name for ds/dt and d²s/dt²?

velocity and acceleration?

 

[nvn]

That's correct. I mean, can you think of the variable name for these quantities, in post 40?

 

[vanquish]

v_t and a_t

 

[nvn]

That's correct. Want to try to solve the problem now?

 

[vanquish]

well if a_t=0At=0
An=v^2/ρ=75^2/344.6=16.32

Ax=0
Ay=16.14

 

[nvn]

That is correct for a_t and a_n. That is correct for ay, except for the sign. But ax is currently wrong. Try again.

 

[vanquish]

if at=0 then when i take the pojection of At along Ax i get
Ax=At*cos(8.5)

 

[nvn]

That's right; but don't forget to also project a_n onto the x axis.

 

[vanquish]

That's right; but don't forget to also project a_n onto the x axis.

oh wow

Ax=At*cos(8.5)+An*sin(8.5)=-2.41 m/s²
Ay=-16.14 m/s²

EDIT: -2.41

 

[nvn]

Always maintain four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits, unless the first significant digit of the final answer is 1, in which case round the final answer to four significant digits. You rounded your angle value too much. Try that one more time.

 

[vanquish]

I'll keep that in mind, but it accepted the answer. Thanks for sticking with me, I can be slow sometimes, I really appreciate it and I do think I learned from this, which is always a good thing.

 

[nvn]

Nice work, vanquish. The correct answer is ax = -2.42 ft/s^2, and ay = -16.14 ft/s^2. You were close in your numeric values. And you have the correct signs. But your units are currently incorrect. They did that on http://en.wikipedia.org/wiki/Mars_Climate_Orbiter#The_metric.2Fimperial_mix-up" and lost 328 million USD in a matter of minutes, when it crashed into Mars.