physics...?

Question

A car traveling at 100 km/hr, stops in a distance of 70 m with constant acceleration. What is the magnitude of the acceleration? in meters per second squared

Answer 1

2 a d = v^2 - vo^2

a is the acceleration (what you're looking for)
d is the distance ( 70 m )
v is the final velocity ( zero )
vo is the initial velocity ( 100 km / hr, which must be converted into meters per second )

Convert the velocity to the appropriate units and solve for the acceleration a.

Answer 2

a = (vf^2 - vi^2) / 2d

(acceleration equals the final velocity squared minus the initial velocity squared divided by 2 times the distance traveled)

a = (0^2 - 27.78^2) / 2(70)
(note: since velocity is in km/hr, and distance is in meters, you need to convert the velocity to meters per second by doing the following:
100 km/hr * 1000 / 3600 = 27.78 m/s)

a = (-771.6) / (140)
a = -5.5 m/s squared

Answer 3

Here is the long way: the distance traveled at time t is d=v*t+a*t*t, where v and a are unknown. You know that d'(0) = 100km/hr, and that when d(s)=70m, d'(s)=0, that gives you 3 equations in 3 unknowns, a, b and s, and from there you crank up the algebra. The non-linearity of these equations is a bluff, really.

If you want a quick answer, the first answer is faster.