bigzee20 said:Homework Statement
A speeder passes a parked police car at a constant speed of 27.6 m/s. At the instant, the police car starts from rest with a uniform acceleration of 2.27 m/s^2. How much t passes before the speeder is overtaken by the police car.
I tried solving for t using t=Vf-Vi /a and i get 12.15sec, which is wrong what am i doing wrong?
bigzee20 said:OMMMMG I love u I GOT IT!
x=1/2a(t)^2 + Vi(t)
Cops= 1/2(2.27)(12.15)+0(12.15) = 13.179025
Speeder= 1/2(0)12.15+27.6(12.15) = 335.34
335.34/13.179025 = 24.32s
The formula for calculating time and velocity is time = distance/velocity. This means that the time it takes for the police car to overtake the speeder is equal to the distance between them divided by the difference in their velocities.
The speed of the police car directly affects the time it takes to overtake the speeder. The faster the police car is traveling, the shorter the distance it needs to cover to catch up to the speeder, resulting in a shorter amount of time to overtake.
Yes, the relative velocity of the police car and the speeder can affect the outcome of the overtake. If the speeder is also increasing their speed, it may take longer for the police car to catch up to them, resulting in a longer time to overtake. On the other hand, if the speeder is decreasing their speed, it may take less time for the police car to overtake them.
Yes, it is possible for the police car to overtake the speeder even if they have the same velocity. This can occur if the police car starts from a position ahead of the speeder or if the speeder slows down, allowing the police car to catch up and overtake them.
Yes, there are other factors that can affect the time and velocity of a police car overtaking a speeder. These include the distance between the police car and the speeder, the acceleration of both vehicles, and any external factors such as road conditions or traffic. These factors can all impact the outcome of the overtake and should be taken into consideration when calculating time and velocity.