Chasing crims

You have taken a part time job as the technical assistant to the director of an adventure film to be made here in Edinburgh. The script calls for the hero to apprehend fleeing criminals by jumping onto their car. The stunt is due to be filmed on one of the roads around Edinburgh, which has a 1.2km long straight section.

The director has asked you to determine if the shot can be completed according to the sctipt, and if the straight bit of road is long enough to execute the stunt safely. The director is paying £ 10K an hour for the helicopter that will film the action, so you need to get this right.

In the script, the bad guys come around a corner onto the straight stretch at 95 km/h, passing the hero’s car that is concealed and stationary at the start of the straight stretch. It takes the hero 3 seconds to start the engine, floor the accelerator and take off with a constant acceleration of 2 ms-2.

Hide

Hint

You should be getting used to applying the FDPEE strategy to problems like this by now.... the key phyiscs here is the kinematics equations at constant acceleration.
 

Solution

Reveal
Hide

Solution

Describe the Physics:

Well call the two cars B for badguys and H for hero. We’ll also identify 3 key parts of the motion.

1. Badguy appears and passes stationary hero

2. Hero takes off

3. Hero catches crims.

Event (3) will happen when the displacement of the two cars from the origin (let’s say that is where the Hero takes off from) is equal. Let’s also denote the time t as the time elapsed since the hero took off. (Note that the time for the badguys will be t+3).

Then we can write expressions for the displacements of each of the vehicles as a function of time, and equate:

$$x_H = v_{0H}t + \frac{1}{2} a_H t^2 = 0 + \frac{1}{2} a_H t^2$$

and

$$x_B = v_{0B}(t+3) + \frac{1}{2} a_B (t+3)^2 = v_{0B}(t+3) + 0$$

Equating, we get

$$\frac{1}{2} a_H t^2 = v_{0B}(t+3)$$

Which rearranges to a quadratic in t:

aHt2-2v0Bt-6v0B=0

We need to convert the speed from kmh into m/s, and solve: only the positive value for t from the two solutions makes sense here and this is calculated to be 29s.

Either of the two expressions for the distance travelled gives us that

x=850m

So, the stunt will probably work on this stretch of road.