Bad Physics in a Movie:

SPEED (1994)Project by: Esteban Lee, Vicky Liu, Brenda Yang , David Liu

Let’s look at that again:

What just happened? The passenger bus left the end of the freeway at an incline,

flying over the gap for a duration of 7.33 seconds, and then landed on the other end of the freeway.

The time was determined by timing the start and end of the bus’s projectile motion in the scene.

We can make the assumption that the two ends of the freeway are level, since it was a gap in an already existing freeway, and it looks very equal in the video.

HEIGHTS ARE

EQUAL= =

bridge bridgegap

But what’s the length of the gap?

50 FEET

50 ft50 ft x 0.3048= 15.2 meters

From that clip, we determined the length of the gap as 50 ft, which is 15.2 meters when converted into meters.

θ = 20.5°

θ

N

THE BUS LEAVES THE RAMP WITH AN ANGLE OF 20.5° up from the horizontal.

Next, we found the angle of the projectile:

THE VELOCITY OF THE BUS In the movie, the speedometer displays a velocity of 67 mph right before the bus

leaves the end of the highway. When the velocity is converted, you get:

Therefore, with the angle, the initial velocity the bus leaves the ramp is 30 m/s, 20.5° from the horizontal.

67 mp/h67m x 1.60934= 107.8258 km/h

107.8258km/h ÷ 3.6 = 29.95 m/s ≈ 3.0 x 10^1 m/s

Here is what we have so far:

dx = 15.2 meters

Force of Gravity

20.5°Vo = 30.0m/s

time: 7.33 seconds

Acceleration (y) : -9.8m/s2

Acceleration (x): 0 m/s2

dy= 0 meters

Projectile MotionO In the clip, we saw the bus leave the bridge

at an angle of 20.5° relative to the bridge. We assumed that there was a reason why the bus left the bridge at such an angle. After calculating the horizontal displacement, we proved that even at an angle of 20.5°, the bus would still have fallen off the bridge.

What we know…

O θ=20.5° O a=-9.8 m/s2

O dy=0 m

- Both sides of the bridge were at the same height, so the y displacement is 0

O Vo=67 mph = 29.95 m/s

- The velocity when the bus left the bridgeO Voy=29.95sin20.5°=10.49

m/sO Vox=29.95cos20.5°=28.05

m/s

20.5 °

CalculationO Calculate the time that the bus required to land on

the other side (same elevation) (y displacement = 0). dy=Voyt+1/2at2

0=10.49t+1/2(-9.8)t2

0=10.49t+(-4.9)t2

0=t(10.49t – 4.9) t→10.49t – 4.9 = 0 10.49t = 4.9 t=0.476s and 0sO 0 s is when the bus started to leave the bridge, so

0.476 s is when the bus landed at the same height as it started.

CalculationO Calculate how far the bus would travel in 0.476 s

(x displacement). dx=Vx ∙ t

dx=28.05 ×0.467

=13.09935m ≈ 42.98ft → 43ft O With an angle of 20.5°, the bus would have landed

43ft away from the bridge which still could not reach the other side of the bridge, located 50ft away.

Therefore, with the information provided, it would NOT be possible for the bus to clear the gap.

HOWEVER, there was one flaw. If you look closely in the movie, you can clearly see no distinguishable ramp or incline at the end of the bridge.

WHERE’S THE RAMP OR INCLINE??

STILL NOTHING HERE.

Another photo that shows no ramp or incline

THERE IS NO RAMP OR INCLINE.

**This is the side rail of the freeway, not a ramp.

Since there was no ramp or incline to physically give the bus a lift, the bus should have simply fallen off the bridge.

THE ‘Y’ COMPONENT

dy= ?2

dy = Voyt + (1/2)at²dy = (0)(7.33s) + (1/2)(-9.8m/s)(7.33s)²dy = 0 + (-4.9)(53.7289)dy = -263.27161mdy = -863.752 ft

dy ≈ -860ft  

Calculating the ‘y’ displacement

Based on what was seen in the video, after 7.33 seconds have passed, the bus should have fallen 860ft below the height of the bridge.

THE ‘X’ COMPONENT

dx= ?

Calculating the ‘x’ displacement

dx = (vx)(t)dx = (29.952m/s)(7.33s) dx = 219.54816mdx = 720.295ft

dx ≈ 720ft 

Based on what was seen in the video, after 7.33 seconds have passed, the bus should have travelled a total x distance of 720ft. 

dx= 720 ft

dy= -860 ft

Without the ramp, the bus would have landed where the smiley face is, instead of clearing the gap.

In order for the bus to actually make the leap and land fi fty feet away (same elevation), various adjustments must be made. First of all, there must be a ramp with a sharp incline. Moreover, the bus’s initial velocity must be considerably greater.

CAN THIS SITUATION BE MADE PHYSICALLY POSSIBLE?

Now find the Y-component of the bus’s initial velocity in this ideal situation, where it jumps and lands perfectly in the given time range of 7.33 seconds.

With significant digits, the bus must leave the bridge at 4.0 x 10^1 m/s, at 87 degrees from the horizontal, in order to successfully complete the jump. Compare this to the bus’s speedometer reading of 67 miles per hour, or 29.95 m/s.

Fg = mgFg = (10570kg)(9.8m/s^2)Fg = 103586 N 1.0 x 10^5 N

The bus would still sustain serious injuries!

Movie : Speed(1994) Bus Model: GM New Look bus Length: 29ft (8.84m) Width: 96 in(2.24m) Height: 121 in(3.07m) Mass: 23,310 lb(10,570kg)

Limits and Assumptions

• We assumed that the height of the bridge was equal before and after the gap.

• We assumed that the gap was exactly 50ft, so we took it as a count.

• We assumed there was no friction or air resistance.

• We assumed the angle was 20.5 degrees by measuring it with a protractor, since it had a physically impossible magnitude anyway.

• Esteban assumed Mr. Ahn would be okay with the swearing in the video.

T H A N K YO U F O R L I S T E N I N G .

LA FIN