Basic problem on the equation of motion

[Frisky90]

Homework Statement

The brakes on a car are applied when traveling at 55km/h and it comes to rest uniformly in a distance of 50m. The mass of the car is 800 kg . Calculate the braking force and the time for the car to come to rest.

Homework Equations

F=ma
v=u+at
t=s/v=50*3.6/55=3.27s

0=55/3.6 +a*3.27
a=-4.67 deceleration. Then F=800*4.67=3738.2263N braking force
However, according to the txtbook the right answers are: 1.87kN, 6.55s ?

 

[Apphysicist]

Homework Statement

The brakes on a car are applied when traveling at 55km/h and it comes to rest uniformly in a distance of 50m. The mass of the car is 800 kg . Calculate the braking force and the time for the car to come to rest.

Homework Equations

F=ma
v=u+at
t=s/v=50*3.6/55=3.27s

0=55/3.6 +a*3.27
a=-4.67 deceleration. Then F=800*4.67=3738.2263N braking force
However, according to the txtbook the right answers are: 1.87kN, 6.55s ?

Watch your units...your kinematics equations deal with m/s and m/s/s. You have km/h.

Also, from an energy standpoint, I verified that 1.87kN is correct for the force.

 

[Frisky90]

Watch your units...your kinematics equations deal with m/s and m/s/s. You have km/h.

Also, from an energy standpoint, I verified that 1.87kN is correct for the force.

I can't find a problem with the units.I've converted km/h to m/s.
And I know that the textbook is correct :)

 

[Chronos]

You have used distance traveled inappropriately. The correct way to plug in distance is
a = (v^2 - u^2) / 2s
where v is final velocity, u is initial velocity, and s is distance traveled.

 

[Frisky90]

You have used distance traveled inappropriately. The correct way to plug in distance is
a = (v^2 - u^2) / 2s
where v is final velocity, u is intial velocity, and s is distance traveled.

I don't understand how do you derive this expression for a? Could you explain? And if you plug in the numbers in it, you still don't get the correct answer.

 

[Chronos]

Permit me to clarify. Distance traveled is the area under the line of the time - velocity graph. Since this is a straight line when acceleration is constant, the area under the line is a simple triangle, so the formula is
s = [(v + u)*t] / 2 or
t = 2s / (v + u)

for s =50, v = 0 and u = 15.273

t = 100/15.273 = 6.548 seconds.

solving for [a] we use v = u + at and get

0 = 15.273 + at
0 = 15.273 + 6.548a
6.548a = -15.273
a = -2.332

The results are the same using the expression I originally gave, albeit that one is a little harder to derive.