Question

The stopping distance of a vehicle is an important safety feature. Assuming a constant braking force is applied, use the work-energy theorem to show that a vehicles stopping distance is proportional to the square of its initial speed. If an automobile traveling at 45 km/h is brought to a stop in 50 m, what would be the stopping distance for an initial speed of 90 km/h?

Answer 1

Answer:

The stopping distance would be 200 m.

Explanation:

Hi there!

The work done to stop the vehicle is equal to its change in kinetic energy.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass.

v = velocity.

The change in kinetic energy is calculated as follows:

ΔKE = final kinetic energy - initial kinetic energy

In this case, the vehicle is brought to stop, so, the final kinetic energy will be zero.

ΔKE = 0 - 1/2 · m · v²

The work done is calculated as follows:

W = F · d

Where:

W = work done

F = applied force

d = traveled distance (stopping distance in this case)

The force F is calculated as follows:

F = m · a

Where:

m = mass

a = acceleration

Then:

W = ΔKE

F · d = -1/2 · m · v²

m · a · d = -1/2 · m · v²

a · d = -1/2 · v²

d = -1/2 · v² / a

Let´s find the acceleration of the vehicle that is brought to stop in 50 m with an initial velocity of 45 km/h.

Let´s convert 45 and 90 km/h into m/s

45 km/h · 1000 m/ 1 km · 1 h /3600 s = 12.5 m/s

90 km/h · 1000 m/ 1 km · 1 h /3600 s = 25 m/s

The distance and velocity of the vehicle is calculated using the following equations:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = traveled distance at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity at time t.

Let´s place the origin of the frame of reference at the point where the vehicle begins to decelerate so that x0 = 0. When the vehicle stops, its velocity is zero. Let´s use the equation of velocity to find the time it takes the vehicle to stop (and travel a distance of 50 m):

v = v0 + a · t

0 = 12.5 m/s + a · t

-12.5 m/s / t = a

Using the equation of traveled distance, let´s find the time it takes the vehicle to travel 50 m until stop:

x = x0 + v0 · t + 1/2 · a · t²

Replacing a = -12.5 m/s / t

50 m = 12.5 m/s · t + 1/2 · (-12.5 m/s/t) · t²

50 m = 12.5 m/s · t - 6.25 m/s · t

50 m = 6.25 m/s · t

50 m/ 6.25 m/s = t

t = 8.0 s.

Then, the acceleration is the following:

-12.5 m/s / t = a

-12.5 m/s / 8 s = a

a = -1.5625 m/s²

Then, the stopping distance of the vehicle if it travels at an initial speed of 90 km/h would be the following:

d = -1/2 · v² / a

d = -1/2 ·(25 m/s)² / -1.5625 m/s²

d = 200 m

The stopping distance would be 200 m.