A car
speeding at
30km/hr passes a
police car at rest
just at the
speeding
motorist passes
the police car
manintains a
constant of 5.8
m/s^2,when and
where will the
speeding
motorist be over
taken?
30 km/hr is hardly speeding where I live. Not even in a school zone.
30 km/h = 8.333 m/s
Solve 8.333 t = 2.9 t^2 for t. Then use that t tigether with
X = 2.9 t^2 for the location, X.
12.90
To answer this question, we need to analyze the motion of both vehicles and find the point where they meet. We can start by converting the car's speed from km/hr to m/s, since the acceleration is given in m/s^2.
1. Convert the car's speed from km/hr to m/s:
- Speed of the car = 30 km/hr
- 1 km = 1000 m
- 1 hr = 3600 s
- Speed of the car = (30 * 1000) / 3600 = 8.33 m/s (approximately)
Now, let's solve for when and where the speeding motorist will be overtaken by the police car:
2. Set up the equation of motion for the speeding motorist:
- Initial speed of the speeding motorist (u) = 8.33 m/s (since they are passing the police car at rest)
- Acceleration of the speeding motorist (a) = 5.8 m/s^2
- Distance covered by the speeding motorist (s)
- Time taken by the speeding motorist (t)
Using the equation of motion: s = ut + (1/2)at^2
3. Set up the equation of motion for the police car:
- Initial speed of the police car (u) = 0 m/s (since it is at rest)
- Acceleration of the police car (a) = 0 m/s^2 (since it is not moving)
- Distance covered by the police car (s)
- Time taken by the police car (t)
Using the equation of motion: s = ut + (1/2)at^2
Since the police car is not moving, its equation becomes: s = ut
4. Equate the distances covered by both vehicles:
- Distance covered by the speeding motorist = Distance covered by the police car
ut + (1/2)at^2 = ut
5. Simplify the equation:
(1/2)at^2 = 0
Since the only way for the equation to hold is if t = 0, this means that the speeding motorist will overtake the police car at the exact moment they pass the police car.
6. Determine the location of the overtaking:
To find the location of the overtaking, we need to substitute the value of t into one of the equations of motion.
Using the equation: s = ut + (1/2)at^2
Distance covered by the speeding motorist = u * t + (1/2) * a * t^2
Distance covered by the speeding motorist = 8.33 * 0 + (1/2) * 5.8 * (0)^2
Distance covered by the speeding motorist = 0
Therefore, the speeding motorist will be overtaken at the exact location they pass the police car.
In summary, the speeding motorist will be overtaken by the police car at the same moment they pass the police car, and it will happen at the same location.