Expert: Jack Cheng - 11/14/2006

Question
Hi Jack, I am having alot of problems with this question and I would be very grateful if you could help. It is from a task we have in VCE General Mathematics 2. I don't expect you to do the whole lot for me, I tihnk if you could explain the first question A I could do the rest, and some hints would be good too! Also I can't do the last question as I have never understood how to draw a velocity time graph properly. My teacher has said that you have to get things in terms of (t) first, and draw a graph and use the areas to work out the time as the time is the same for both so one area can equal the other. Hope I haven't confised the issue, if so just ignore those and do it any way! Thank You so much. From Belinda.


QUESTION...
A police officer is sitting at the side of the road on a motorbike with the engine running, monitoring motorist's speeds with a speed gun. The speedlimit is 100 km/h. A motorist passes the police officer at 120 km/h and as the car passes, the police officer starts from rest and chases the motorist with uniform accelaration of 3 m/s^2.
A) How long does it take for the police officer to reach the car?
B) What is the maximum speed of the police offficer?
C) What is the distance travvled by the two vehilcles, when the police officer reaches the car.
D) What distance does it take for the car to come to a stop?
E) Skatch a velocity time graph of the motion of the two vehicles on the same set of axes.  

Answer
Hi Belinda,

For this question, you have understand the relationship among distance, velocity, and acceleration. The acceleration is most likely constant (you have to go into calculus otherwise). To keep this short, here are the formulas:

vf = vi + at     
d = vi*t + 0.5at^2
vf^2 = vi^2 + 2ad

(d = displacement/distance; a = acceleration; vf = final velocity; vi = initial velocity; t = time)

A) For the speeder, his velocity is 120 km/h and his acceleartion is 0. So you can use the second equation to find the distance.

For the police, you do the same thing with equation two, except "vi" is now 0 (at rest) and "a" is 3 m/s^2 (be careful of unit conversions, 120 for the speeder is in km/h). Now, you set the two equations equaling to each together and solve for t.

B) For the maximum speed of the police officer, you can use first equation above. t is what you got from A).

C) Similar to A), but since you know the time, you can now solve for the distances.

D) I am not understanding this question. The scenario mentions nothing about the cars slowing down.

E) Graphing. Okay, for the speeder, his velocity is constant, so it's just a horizontal line.

For the officer, though, it's a bit tougher. At time t=0, the officer is not moving. So v=0. Then, at each t, the velocity increases at a constant rate of 3 m/s^2; in other words, the velocity increases by 3 every second. This is a linear graph, with slope=3 and y-intercept=0.

I hope this is enough to get you started. If not, ask again and I'll show you more details.

~ Jack