Question

CALCULUS HELP

An Acura NSX going at 70 mph stops in 157 feet. Find
the acceleration, assuming it is constant.
I'm kind of confused, could I get a bit of help to push me towards the answer?

Answer 1

assume it goes form 0 to 70 in 157 feet ... might be simpler

Answer 2

and convertion from feet to miles might be useful

Answer 3

I can solve this using the equations for uniform acceleration which are derived using calculus

u = 70
v = 0
s = 157
a = ?

v^2 = u^2 + 2as

0 = 70^2 + 314 a
a = -(70^2) / 314

Answer 4

I didn't want to use the kinetic equations, but I might have to.

Answer 5

Sorry, kinematic

Answer 6

let a = a

v = at + k, v(0) = 0 soo

v = at

d = 1/2 at^2 + c, but d(0) = 0 sooo

157feet = 1/2 at^2

sqrt(2(157feet)/a) = t

70mph = a sqrt(2(157feet)/a)

70mph = sqrt(2a 157feet)

Answer 7

i've forgotten the derivation i'm afraid
now doubt Amistre will be able to help

Answer 8

Ok, thanks. That makes sense.

Answer 9

\[\frac{dv}{dt} = -a\]
does that mean anything to you?

Answer 10

70mph^2 = 2a 157feet

70mph^2
--------- = a for whatever thats worth lol
2 *157feet

Answer 11

I've already solved the problem...

Answer 12

for a more general feel
\[d_1=\frac12 at^2\]
\[\sqrt{\frac{2d_1}{a}}=t\]
\[v_f=a\sqrt{\frac{2d_1}{a}}\]
\[v_f=\sqrt{2d_1~a}\]
\[\frac{v_f^2}{2d}=a\]