I need help in Basic Physics?

Question

A sports car is advertised to be able to stop in a distance of 55 m from a speed of 105 km/h.

(a) What is its acceleration in m/s2?

______m/s2

(b) How many g's is this (g = 9.80 m/s2)?

_______

ANOTHER ONE

A car slows down from a speed of 27.0 m/s to rest in 6.00 s. How far did it travel in that time?

_________ m

WRITE STEPS

Answer 1

You have a distance (55 m), an initial speed (105 km/h) and a final speed (0).  You're looking for acceleration.

First off, get everything into consistent units.  Since you want the acceleration in m/s^2, it seems reasonable to turn everything into meters and seconds.  So:

105 km/h * 1 h/3600 s * 1000 m/1 km = 29.1666... m/s

Now:

a = (vf - v0)/t

s = (vf + v0)t/2 which give us t = 2s/(vf + v0)

Therefore,

a = (vf - v0)/(2s/(vf + v0)) = (vf^2 - v0^2)/2s

= [0^2 - (29.1666... m/s)^2)/2(55m) = -7.336 m/s^2

I'll let you worry about significant digits and where that minus sign came from.

This is 7.336/9.80 = 0.789 g's

For the second:

Use the equation we derived above.  Watch your signs and your units.