A speeding motorist passes a stopped police car. How fast is the speeding car going?
I need this explained to me, not just answered.
A speeding motorist passes a stopped police car. At the moment the car passes, the police car starts from rest with a constant acceleration of 4.28 m/s2. The speeding motorist continues with constant velocity until caught by the police car14.8s later. How fast is the speeding car going?
Thank you, Druisse! My main problem is really just.. understanding the problem! Ha ha! Your help is greatly appreciated. To the two who said he wasn’t fast enough.. thank you, too! I thought it funny. (=
The answer is apparently 31.7 m/s. Why should it be divided by 2?
When the police car catches the speeding car they certainly do not have the same velocity. The police car catches up with the speeding car because it has a higher velocity than the speeding car. What is true is : They have covered the same distance, and/ or been travelling for the same time.
What distance did the police go travel?
s = ut + 0.5*a*t²
s = 0 + 0.5*4.28*14.8²
s = 468.75m
The speeding car traveled 468.75m in 14.8 sec,
Velocity of car = 468.75/14.8 = 31.68 m/s
Answer: 31.7m/s
May 23rd, 2010 at 9:59 pm
not fast enough. he got caught
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May 23rd, 2010 at 10:07 pm
Not fast enough if the cop caught him.
References :
lol
May 23rd, 2010 at 10:39 pm
integral of 4.28x dx , from 0 to 14.8
that integral gives the distance covered.
divide the distance by the time and you’ll get the constant speed of the speeding car.
integral of 4.28x dx from 0 to 14.8 = 468.7 m
468.7 / 14.8 = 31.67 m/s
the answer is not 14.8 * 4.28 = 63.34
if one car is going that speed the entire time, and the other one only reaches that speed at the end of accelerating, they will obviously have not gone the same distance. The cop must be going faster than the speeding car to catch him..
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May 23rd, 2010 at 10:47 pm
note: when the police car catches the speeder, they have the same speed or velocity.
a = acceleration
vf = final velocity
vi = initial velocity = 0
t = time
a = (vf – vi)/t
4.28 m/s^2 = (vf – 0)/ (14.8 s) let me remove the units for ease of presentation
(4.28)( 14.8) = vf
63.34 m/s —-> speed of police car at the time it catches the sppeding car =======> and is equal to speed of speeding car
Answer: 63.34 m/s
References :
May 23rd, 2010 at 11:28 pm
When the police car catches the speeding car they certainly do not have the same velocity. The police car catches up with the speeding car because it has a higher velocity than the speeding car. What is true is : They have covered the same distance, and/ or been travelling for the same time.
What distance did the police go travel?
s = ut + 0.5*a*t²
s = 0 + 0.5*4.28*14.8²
s = 468.75m
The speeding car traveled 468.75m in 14.8 sec,
Velocity of car = 468.75/14.8 = 31.68 m/s
Answer: 31.7m/s
References :