SITEMAP   School-college Physics Notes: Forces & motion 5.8 Advanced braking calculations

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Forces and Motion: 5.8 Some advanced calculations on braking force and kinetic energy using formulae for force, acceleration & KE

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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/s2)

W = work done (J)

F = force (N)

d = distance (m)

Formulae required

W or KE = 1/2mv2, F = ma, W = Fd, s = d/t, v2 - u2 = 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 v2 - u2 = 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

v2 - u2 = 2ad, 0 - 182 = 2 x a x 30

60a = -324, therefore a = -324/60 = -5.4 m/s2 (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 mv2 = ½ x 1000 x 26.842 3.6 x 105 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 mv2 = ½ x 43,000 x 26.8421.55 x 108 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/s2

(ii) F = ma from Newton's 2nd Law, force in newtons, mass in kg, deceleration in metres per second2

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 = ½mv2 = 0.5 x 1500 x 302 = 675000 = 6.75 x 105 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 )

W = F x d = KE = ½mv2 = 6.75 x 105 J

W = F x d,  d = W / F = 6.75 x 105 / 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 = ½mv2 = 0.5 x 1500 x 182 = 243 000 = 2.43 x 105 J

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

2.43 x 105 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/s2

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

From Newton's 2nd Law: F (N) = m (kg) x a (m/s2)

Decelerating force = 1500 x -90 = 135 000 = -1.35 x 105 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 mv2 = 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 mv2,  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

Keywords, phrases and learning objectives for the physics of road vehicles - advanced 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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