**
5.8 Some advanced calculations on braking force and kinetic energy**

**Diagram KEY**:

**KE** = kinetic energy (**J**),
** **

**m** = mass (**kg**)

**u** = initial velocity (**m/s**),

** v** = final velocity (**m/s**)

**s** = speed (**m/s**)

**a** = acceleration or
deceleration (**m/s**^{2})

**W** =
work done (**J**)

**F** = force (**N**)

**d** = distance (**m**)

**
Formulae required**

W or **
KE = 1/2mv**^{2},
**F = ma**,
W = Fd,
s = d/t,
**v**^{2}
- u^{2} = 2ad

**Q1**
Suppose a car of 1200 kg is travelling at 18 m/s (~40 mph) has to do an
emergency stop with a hazard 30 m ahead.

(a) Calculate the **deceleration of
the car** and (b) the **braking force** involved.

(a) First use the motion equation
**v**^{2} - u^{2}
= 2ad to calculate the deceleration.

where v = final velocity, u = initial velocity, a =
acceleration (∆v/∆t), d = distance travelled

Assuming uniform deceleration and v = 0 (comes to
halt), u = 18 m/s, d = 30 m

v^{2} - u^{2} = 2ad, 0 - 18^{2}
= 2 x a x 30

60a = -324, therefore a = -324/60 =
**
-5.4 m/s**^{2}
(note the negative sign for deceleration)

(This is easier to do if you are
given the braking time, so can just use a = ∆v / ∆t, which
I have done in Q2 next)

(b) You then use Newton's 2nd Law equation
**F = ma**,

where F = decelerating braking force, m = mass of car,

a =
deceleration of car = change in speed / time taken

Substituting into the equation (and you can ignore
the -ve acceleration sign here, but NOT above)

F = ma = 1200 x 5.4 = **
6480 N**

Comment: That's why your body is thrown forward. The
deceleration is just over half the value of the acceleration you
experience due to the Earth's gravitational field. If you are
involved in a high speed impact the force can be much greater and
hence destructive on you and the car!

**See section on**
safety features of road transport

**
Q2 ** A small domestic 1000 kg car (1 tonne)
with two axles at 60 mph (26.84 m/s)

will have a kinetic energy = 0.5 mv^{2} = ½ x 1000
x 26.84^{2} = **
3.6 x 10**^{5} J (360 kJ, 3 s.f.)

A heavy articulated goods vehicle of 6
axles may weigh, with a full load, up to 43 000 kg (43 tonnes) at 60 mph
(26.84 m/s)

will have a kinetic energy = 0.5 mv^{2} = ½ x
43,000 x 26.84^{2} =
**1.55 x 10**^{8} J
(15 500 kJ, 3 s.f.)

Now, both of these vehicles have to be
able to stop in the same safe distance in an emergency.

The two axle car will have four sets
of brake pads.

The six axle goods vehicle will have
twelve sets of brake pads, three times as many as the car.

This means to stop in the same safety
distance, the braking force exerted by each set of pads in the goods
vehicle must be much greater than for the car.

At 50 mph (22.37 m/s) suppose the
safe braking distance is 38 m.

We can then calculate the total
braking force needed to stop in three seconds.

(i) for both vehicles deceleration **
a = ****∆v / ∆t** = 22.37 / 3 = **
7.457 m/s**^{2}

(ii) **F = ma**
from Newton's 2nd Law, force in newtons, mass in kg, deceleration in
metres per second^{2}

For the car: F
= 1000 x 7.457 = **7 460 N**
(3 s.f.),

that is
1865 N braking force per set of four brake pads.

For goods
vehicle: F = 43 000 x 7.457 =
321 000 N
(3 s.f.).

**
that is
26750 N braking force per set of brake pads.**

This means the
heavy goods vehicle brake pads must generate over 14 x the braking force
of the car.

(For those expert
in road vehicle physics, I do appreciate these are simplified
calculations)

For more on F = ma calculations
see
Newton's Second Law of
Motion and
momentum calculations

Q3 Suppose
a car travelling at 30 m/s (~70 mph) has to make an emergency stop to avoid
a hazard.

If the mass of the car is 1500 kg, the
braking force of the car is 6000 N and the **tired** driver's
reaction time is 1.5 seconds, calculate the following:

(a) **Calculate the thinking
distance of the driver** (s = speed (m/s), d - distance (m), t = time
(s))

s = d / t, d = s x t = 30 x 1.5 =
**45 m = thinking
distance**

(b) **Calculate the initial kinetic
energy of the car** (m = mass of car in kg, v = speed of car (m/s)

KE = ½mv^{2} = 0.5 x 1500
x 30^{2} = 675000 = **
6.75 x 10**^{5} J =
initial KE of car

(c) **Calculate the braking distance
to halt the car** (W = braking work done (J), d = braking
distance (m)

Work done in braking the car must
equal the kinetic energy of the car (see
Graph 3 part 5.3)discussion)

W = F x d = KE = ½mv^{2}
= 6.75 x 10^{5} J

W = F x d, d = W / F = 6.75
x 10^{5} / 6000 =
**113 m = braking distance** (3 s.f.)

(d) Calculate the stopping distance
of the car

stopping distance = thinking
distance + braking distance

= 45 + 113 = **
158 m = stopping distance**

**
Q4** **See**
reaction
time experiment

**
Q5** A 1500 kg 4x4 car travelling at 18.0
m/s (~40 mph) veers off the road, without slowing until hitting and
demolishing a brick wall.

If it took 0.200 seconds to demolish
the wall, **calculate the following** ...

(a) **What is the initial kinetic
energy of the car?**

KE = ½mv^{2} = 0.5 x 1500
x 18^{2} = 243 000 = **
2.43 x 10**^{5}
J

(b) **What work is done on the wall
and car in bringing the car to a halt?**

**
2.43 x 10**^{5}
J because all the kinetic energy of the car has to be
removed.

(c) **What happens to the kinetic
energy of the car after impact?**

**The kinetic energy store of
the car is reduced to zero** and the energy is converted into heat
(by compression or friction) and some sound energy (which will end
up as heat too). **So the thermal energy store of the wall, car and
surrounding air is increased**.

(d) **Calculate the rate of
deceleration**

Deceleration = change in speed /
time taken = ∆v / ∆t = (0 - 18) / 0.2 = **
-90 m/s**^{2}

(e) **What is the
decelerating force acting on the car?**

From Newton's
2nd Law: **F **(N) **= m **(kg)** x a **(m/s^{2})

Decelerating
force = 1500 x -90 = 135 000 =
**
-1.35 x 10**^{5}
N

The force
(from the wall) is negative because it is acting in the opposite
direction to the motion of the car.

If the car had
braked in time, the decelerating force would be positive (in every
sense of the word!).

**
Q6** Imagine a car of 1000
kg travelling at 20 m/s doing an emergency stop in a distance of 25 m - the
braking distance.

Calculate the average braking force
produced by the driver when pressing on the brake pedal.

To solve this question you to use
several formulae.

(a) Calculate the kinetic energy of
the car.

KE = 0.5 mv^{2} = 0.5 x
1000 x 202 = **200 000 J**

(b) What work must be done to bring
the car to a halt? Explain your answer.

If the kinetic energy of the car
is 200 000 J, then
200 000 J
of work must be done to bring the KE of the car to zero i.e
zero velocity.

(c) Calculate the average braking
force required.

Work (J) = force (N) x distance
(m)

work = 200 000 J and the braking
distance was 25 m

force = work / distance = 200 000
/ 25 = **
8000 N average braking force.**

**
Q7** A van of mass
2000 kg veers off the road at 30 m/s and becomes stationary after hitting a
stone wall.

(a) If the impact force on the van is
48 000 N, calculate the stopping time.

**
F = m∆v
/ ∆t**, substituting

48 000 = 2000
x (30 - 0) / **
∆t**

48 000 = 60
000 / **
∆t**

**
∆t** = 60 000 / 48
000 = **
1.25 s**

(b) Explain how a safety belt and an
inflating air bag can save the drivers life.

On impact, the driver's body is
accelerated forwards.

(i) The safety seat belt
stretches sufficiently to reduce the rate of change of momentum -
increasing the deceleration time.

(ii) The 'soft' inflated airbag
also reduces the rate of change of momentum and absorbs kinetic
energy when the driver's body hits it.

**
Q8** A 20 000 kg road
vehicle comes to an emergency halt.

A uniform braking force of 8000 N
is applied by the driver until the vehicle comes to a halt in a
distance of 20 m.

(a) Calculate the speed of the
vehicle just before the brakes were applied.

Work done in braking =
braking force x distance brakes applied = 8 000 x 20 = 160 000 N

The total work done in
braking = the kinetic energy of the vehicle at the instant the
brakes are first applied.

KE = 0.5 mv^{2},
rearrangement gives v = √{(KE / (0.5 x m)}

v = √{(160 000 / (0.5 x 20
000)} = **4 m/s**

(b) What are the major energy
store transfers taking place?

The kinetic energy of the
vehicle is mainly converted, via friction, to increase the
thermal energy store of parts of the vehicle and the surrounding
air or road.

**For more on kinetic energy
calculations see **
Kinetic
energy store calculations

INDEX of physics notes on
reaction times, stopping distances of road vehicles, Newton's 2nd Law,
braking friction force, KE calculations

**
Keywords, phrases and learning objectives for **
**the physics of road vehicles - a****dvanced braking
force and kinetic energy calculations**

Be able to do advanced calculations on braking force and kinetic energy using
the various formulae
for acceleration and kinetic energy.

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INDEX of physics notes on
reaction times, stopping distances of road vehicles, Newton's 2nd Law,
braking friction force, KE calculations