whats the equation to find acceleration during braking?

Question

The nominal stopping distance of the sports car is 130. ft from 60.0 mi/h. Determine the acceleration in m/s^2.

a. 6.0 m/s^2
b. - 9.08 m/s^2
c. - 6.02 m/s^2
d. 50 m/s^2

Answer 1

formula is Ke=MA2

Do your own homework

Answer 2

convert the figures into SI units.. therefore
130 ft = 39.624m and
60mi/h = 26.67 m/s

then we can use the eqn of motion assuming that all other forces or natural resistances are ignored.. the

v^2 = u^2 + 2as
if we can re-arrange the formula to make "a" the subject then

a = (v^2 - u^2)/2s
hence
v= 0 m/s
u = 26.67 m/s
s = 39.624 m

replacing and thus a= -8.975 m/s^2.

Answer 3

well acceleration in this case is deceleration...

the eqn is the same...

v0 = 60 mph
vf = 0 mph
x = 130 ft = .0246 mi...

so

.0243 = v0*t + .5*a*t^2 = 60*t + .5*a*t^2

where t can be found as per the following

vf = vo + a*t => 0 = 60 + a*t...

t = -60/a...

so,
.0243 = 60*(-60/a)+.5*a*(-60/a)^2 ... solve for a

reminder... the acceleration from that equation will be in mi/hr^2
you will have to convert to m/s^2
i wont do that for you

Answer 4

First, convert everything to MKS.

130 ft. = 39.624 m
60 m/h = 26.82 m/s

Now remember that

v²=v0²+2ax

where v is final velocity, v0 is initial velocity, a is acceleration, and x is distance

0²=26.82²+2a*39.624

and solve for a = -9.076 m/s² (a negative acceleration is decceleration)

so choose answer b

Which says that car has one *helluva* good set of tires. A traction coefficient of 1 (perfect traction, no slippage) would mean a maximum deceleration of -9.8 m/s² (1g)


Doug

Answer 5

you need to use 2 equations: v = at and d = vt+.5at^2 given which you comprehend neither a nor t, you need to precise one equation in terms of t and then replace into the different equation.Then sparkling up for a. t = v/a d = v^2/a +,5a(v^2/a^2) = v^2/a + .5v^2/a = a million.5v^2/a a = a million.5v^2/d

Answer 6

v^2/(2*x)

160^2/(2*.03) = 430000 Mi/hr^2

or ~175 feet/sec^2

or ~56 m/s^2