13th September 2007 - 08:32 AM
QUOTE (meeshee+Sep 13 2007, 01:59 PM)
A speeder passes a parked police car at 30 m/s. The police car starts from rest with a uniform acceleration of 2.44 m/s/s. (a)How much time passes before the speeder is overtaken by the police car? (b) How far does the speeder get before being overtaken by the police car?
(I know this is a simple problem; I just don't know how to do it!)
Note - I'm using Unicode for special characters.
For the police officer:
d = ½ . a . t²
For the speeder:
d = v . t
The police officer can be considered to have passed the speeder when they have travelled an equal distance, thus, we substitute the second equation into the first, giving us.
v . t = ½ . a . t²
Solving this for t gives us:
v . t = ½ . a . t² (÷ t)
v = ½ . a . t (÷ a)
v/a = ½ . t (× 2)
2.v/a = t
Substituting in the values gives us:
t = 2.v/a
t = 2 . 30 / 2.44
t = 24.590 s
Now we know how long it took.
v = d/t
d = v.t
d = 30 . 24.590 s
d = 737.705 m
13th September 2007 - 09:34 PM
There is a nice graphical way to solve this problem, too. Draw the velocity versus time graph for the car and the cop on the same time axis. The car's graph is a horizontal line of height 30, and the cop's graph is a straight line through the origin of height 2.44.
Now the area under a velocity graph shows the distance traveled, and the cop will catch the car when they meet again, so of course they've traveled the same distance. Since the area of a rectangle is bh and of a triangle is bh/2, (b=time, h=velocity) these areas will be the same when the velocity of the cop is exactly 60.
Therefore, the time is t = 60/2.44 and the distance is vt = 30*60/2.44. This gives the same numbers Trippy got.
Hope this helps!
10th March 2012 - 04:42 PM
How would you answer this question:
A police car is at rest, reading the speeds of passing cars. The police car is passed by a car breaking the speed limit at 30 m/s (uniform velocity). It takes 14 seconds later before the police car moves at a constant acceleration of 2.5 m/s2 . How long will it take before the police car catches up with the speeding car?
It's much like the one above only there's 14 seconds at the start to play with.
All help would be much appreciated.
24th March 2012 - 01:26 AM
Same as how you answered the previous question however the speeders equation should be:
This extra addition is due to the 14 seconds of the police car's rest before it begins to move, thus the initial condition of the speeder is different. In the previous question the racer was at d=0 when the cop began its pursuit, however, this question gives the speeder a 14 sec head start, therefore, you must add this head start distance to your speeders equation before you solve.