A speeder traveling at 41 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 3.3 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder?
Im sorry I am new to physics this is my first year and I still have no idea what to do for this problem. The other two that drwls answered I had them correct before drwls answered them and was using it as a check. When I reworked them drwls way, I got the same answer. However on this problem, I am at a loss because I feel there is not enough data, and I am obviously wrong. I worked it differently than u said and got 12.4 seconds, which sounds wrong. Then I worked it again and got 24.8 seconds. Please help!
check your answers:
a. In 12.8 seconds, how far do each travel?
1) d=41*12.4=508.4m
2) d= 1/2 (3.3)12.4^2=253.7 m
well, that time is wrong.
So check the other and see if the distances match.
To solve this problem, you need to use the equation of motion that relates distance, velocity, acceleration, and time:
distance = initial velocity * time + 0.5 * acceleration * time^2
First, let's find the distance that the speeder travels before the policeman starts accelerating. Since the policeman is at rest, the distance traveled by the speeder is:
distance = speed * time
where speed is the speed of the speeder and time is the unknown variable we want to find.
Next, let's find the distance that the policeman travels to catch up to the speeder.
To catch the speeder, the policeman needs to close the distance between them, so the equation becomes:
distance = (speed + acceleration * time) * time
where acceleration is the acceleration of the policeman.
The two distances are equal, so we can set them equal to each other:
speed * time = (speed + acceleration * time) * time
Now, we can solve for time:
speed * time = speed * time + acceleration * time^2
Subtracting speed * time from both sides, we get:
acceleration * time^2 = 0
This quadratic equation has only one solution, which is time = 0.
Thus, the policeman catches the speeder instantly when he starts to accelerate.
So, to the nearest tenth of a second, the time it takes for the policeman to catch the speeder is 0 seconds.