Question

A test of the prototype of a new automobile
shows that the minimum distance for a controlled stop from 94 km/h to zero is 41 m.
Find the acceleration, assuming it to be
constant as a fraction of the free-fall acceleration. The acceleration due to gravity is
9.81 m/s
2
.
Answer in units of g

How much time does the car take to stop?
Answer in units of s

Answer 1

@Agent47 can you help me right here

Answer 2

d=41m
V_i=94km/h = 94,000m/h = 94000m/60s = 1,566.67 m/s
V_f=0
a-?

Answer 3

a=c*g
d=V_i*t+(1/2)(c*g)(t^2)
a=(V_f-V_i)/t

41=1,566.67+(c*g*t^2)/2
c*g=(0-1,566.67)/t

Answer 4

-3051.34 = c*g*t^2
-1566.67 = c*g*t

Answer 5

dividing the two equations gives:
-3051.34/-1566.67=cgt^2/(cgt)
t = 1.95

Answer 6

uhmm
c*g = -1,566.67/t
c*g = -1,566.67/1.95
a = c*g = -803.42

Answer 7

t=1.95s
acceleration is: a = -81.98*g

Answer 8

I got -81.98 from dividing -803.42 by 9.81

Answer 9

okay im writing it all down

Answer 10

this is how its done