Help Me With This Integral Problems?

Question

1. If a car goes from 0 to 80mph in six seconds with constant
acceleration, what is that acceleration.
2. An acura NSX going at 70 moh stops in 157 feet. Find the
acceleration, assuming it is constant.


PLEASE SOLVE THIS PROBLEMS USING THE DEFINITION OF INTEGRAL AND DIFFERENTATION ONLY, NOT USING THE SPEED,ACCELERATION, AND VELOCITY THEORY.

Answer 1

acceleration=rate of change of velocity
=80*5280/3600*6ft/sec^2
=19.6 ft/sec^2

2.0=(70*5280/3600)^2-2f(157)
314s=67.1ft/s^2

Answer 2

1 You know that v = ∫adt
So 80 = ∫[from t = 0 to t = 6] a . dt
So 80 = a∫[from t = 0 to t = 6] dt
80 = a [t] [from t = 0 to t = 6]
80 = 6a
a = 13⅓ miles/h.s (* 5280/60 ft/s² if units are important)
≈1173 ft/s²

2 a = dv/dt
= dv/dx . dx/dt
= v.dv/dx
= d(½v²)/dx
So ½v² = ∫a . dx
½*70² = ∫[from x = 157 to x = 0] a . dx (Because v was 70 to start with (at x = 157 ft, t = 0))

½*70² = a∫[from x = 157 to x = 0] dx
= a (0 - 157)
a = 70 * 53/(-157)
≈ -15.6 miles/h² (* 5280/3600 ft/s² if units are important)
= -22.88 ft/s²

Answer 3

acceleration is the derivative of velocity. (Or, the change in velocity over time.)

a = dv/dt =(v2 - v1)/(t2- t1) = 80/6 = 13.333 mi/(hr x sec)

Assuming the acceleration is constant.
Using the D to be big delta (total change) rather than miniscule.

a=dv/dt -but we don't know t, we do know s (distance =s cuz we're using "d" already).

We also know velocity v=ds/dt = 70mph.

So multiply both sides by 1/v=dt/ds
a (dt/ds) = (dv/dt)( dt/ds) Then cancel the dt's

a(1/v) = dv/ds or a=v(dv/ds)=70*(0-70)/(157-0) = -31.21mphps (approx)

Which is wrong, because I know from Physics that a=v^2/2s, so I've lost a factor of 1/2 in there somewhere.