# Basic problem on the equation of motion

## Homework Statement

The brakes on a car are applied when travelling 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 ?

## Homework Statement

The brakes on a car are applied when travelling 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.

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
Gold Member
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.

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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
Gold Member
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.