# Speed

## The Movie

Speed is a classic 1994 action film where a bus full of civilians has to stay about 50 miles per hour (80 km/hr), or else a bomb planted there by the crazy villain will go off. It’s full of action stunt scenes, some of which are infamous for being highly unlikely or obviously faked.

## Math Situation 1: The Bus Jump

The Speed bus jump is probably one of the most infamous stunts in all movie history. The lead characters Jack (Keanu Reaves) and Annie (Sandra Bullock) divert onto a highway in an effort to keep the bus above 80 km/hr.

But there’s a problem – a 50 foot (15 m) gap in the highway ahead. They’ll have to jump it!

That’s where the problems with this movie start. It’s very unclear exactly what the configuration of the jump is. The bus appears to “magically” pitch upwards as it leaves the first highway, as if there was a jump ramp set up for it.

It’s also unclear exactly where the second piece of highway is in relation to the first – is it at the same level or significantly lower?

Here we’ll analyse two probable scenarios.

**Scenario 1:** The first one will be for the bus not pitching upwards but continuing straight and losing altitude to land on a *lower* section of roadway.

**Scenario 2:** The second scenario will be the “magical” ramp scenario where we assume that the bus manages to pitch upwards at an angle of 30 degrees, but has to land on a section of roadway at the same height.

To keep things simple, we’ll ignore air resistance and the dynamics of the bus rotating in the air.

## Scenario 1 – Flat Jump

The bus not only has to cross the 50 feet of the gap itself, but also another distance equal to the bus itself – the entire bus lands on the far side in the clip, not just the front bit.

A typical city bus might be about 40 feet in length, so the total horizontal distance for the jump is 50 + 40 = 90 ft. Let’s convert that into metres:

90 ft = 90 ft × 0.3048 m / ft

90 ft = 27.43 m

Next up, let’s convert the speed of the bus into km/hr and then m/s:

Bus speed = 50 miles per hour (mph)

Bus speed = 50 miles/hr × 1.609 km/m

Bus speed = 80.45 km/hr

Bus speed = 80.45 km/hr × 1000 m / km / (3600 s / hr)

Bus speed = 22.35 m/s

So now we have the speed of the bus in m/s, and the horizontal jump distance in metres.

To work out how far the bus will fall, we can use the following formula:

*h = 0.5 × a × t ^{2}*

h is the vertical fall distance, a is acceleration (in this case gravity, -9.81 m/s/s) and t is the time duration of the fall

To work out the time duration of the fall, we need to work out how long it will take the bus to cross the horizontal distance:

*jump time duration = distance / speed*

*jump time duration = 27.43 / 22.35*

*jump time duration = 1.227 s*

Now we can substitute this jump duration time *t* into the fall formula:

*h = 0.5 × 9.81 × 1.227 ^{2}*

*h = 7.385 m*

So if the bus lands flat on the far side, it will have fallen a vertical distance of about 7 metres.

## Scenario 2 – Ramp Jump

The bus appears to pitch upwards just as it leaves the first section of highway (as if there’s an invisible ramp there). Let’s calculate what will happen if the bus jumps upwards at an optimal 45 degree angle (ignoring air resistance and other factors):

In this scenario the highway section the bus lands on is *at the same height* as the section it jumps from. We have to find out whether it’s possible to jump the distance.

Once again, we’ll assume that the whole length of the bus crosses over onto the landing section before hitting the ground – so the front of the bus has to travel 90 ft, or 27.43 m.

Now, normally pitching upwards at 45 degrees would take a lot of speed off – but let’s say that under ideal conditions, the bus maintains the 80 km/hr speed as it launches upwards at 45 degrees.

We need to find the *vertical component* of that speed, in order to work out how long that bus will fly upwards, and then downwards before ending back at the same vertical level.

We can use trigonometry, using the *sine *function (**SOH**CAHTOA)

*sin(45) = opposite / hypotenuse*

*sin(45) = vertical speed component / bus speed*

*vertical speed component = bus speed × sin(45)*

*vertical speed component = 80.45 km/hr × sin(45)*

*vertical speed component = 56.89 km/hr*

For the first half of the jump, the bus will be rising. Then it will pause momentarily, with *zero* vertical speed. For the second half, it will be falling.

The rate at which the bus slows in its initial rise will be governed by gravity – 9.81 m/s/s. We can work out how long it will take for the bus’s vertical speed component to drop from 56.89 km/hr to zero at the midpoint of the jump:

*time to reach highest point of jump = initial vertical speed / gravity*

*time to reach highest point of jump = 56.89 km/hr / 9.81 m/s/s*

*time to reach highest point of jump = 56.89 km/hr × 1000 m/km / (3600 s / hr) / 9.81 m/s/s*

*time to reach highest point of jump = 1.61 s*

That makes up the first half of the jump, the *rising* part. There’s also an equally long second falling part. So:

*Total jump duration = 2 × 1.61s*

*Total jump duration = 3.22 s*

So the bus will take 3.22 seconds to rise through the air first, then fall back to its original starting height. Now we can work out how far the bus will travel *horizontally* during this time:

*Horizontal travel distance = horizontal speed component × jump duration*

Now, if you examine the triangle diagram from before, we can see that the horizontal speed component should be exactly the same as the vertical speed component:

*Horizontal travel distance = horizontal speed component × jump duration*

*Horizontal travel distance = 15.80m/s × 3.22 s*

*Horizontal travel distance = 50.88 m*

*50.88 m > 27.43 m*

So in the (very unrealistic) case that the bus pitches upwards at 45 degrees **and** maintains all of its speed, it would theoretically be plausible for the bus to cross the gap and land on an equally high section of highway.

## Real Life Example – Motocross

Motocross involves motorcycle racing on off-road circuits, often over quite challenging terrain. One of the most common challenges is racing at speed over undulating terrain, or stunt shows where they do long jumps, often over cars or other objects using ramps.

Doing these jumps safely and reliably involves making sure that the angles of the ramps, the bike speeds and the jump distances are all correct – a mistake in any of these can result in injuries or worse.