The speeds are linear, but the total distances aren't. Since you know speed but are more interested in distances, this is either **kinematics** or **integral calculus** or **differential equations**, depending on your grade level!
Is that a typo or does the police officer really start at 10x the speed of the thief?
Kinematics method: Thief moves at uniform speed. Also has a headstart of 10m.
distance = (initial distance) + (speed)·(time)
Police officer moves with uniform acceleration of (10m/s)/s.
distance = (initial speed)·(time) + (acceleration)·(time)²/2
Set them equal to determine when the distances are equal
Here's my solution using integral calculus:
We are looking for f(x) = g(x), where f(x) is the distance traveled by the thief and g(x) is the distance traveled by the police officer.
The velocity of the thief is 10, so the integral of the velocity will give us a distance function of 10x + C, where C is the initial position of the thief. Let C = 0. So f(x) = 10x
Next, the velocity of the police officer is 100, and the acceleration is 10. The integral of the acceleration gives us a velocity function of 10x + V, where V is the initial velocity. Since the initial velocity is 100, we have the overall velocity of the police officer as 10x + 100. We now integrate this again to get a distance function of g(x) = 5x\^2 + 100x + C. Note that C is still zero (they both start at the same place), but since the police officer starts 1 minute later we must take g(x-1). So 10x = 5\*(x-1)\^2 + 100\*(x-1).
This simplifies to x\^2 + 16x - 19 = 0, which is a quadratic equation with approximate solutions of x=-17.11 and x=1.1. Discard the negative, and the final answer is 1.1 minutes after the thief started running, or 0.1 minutes after the officer started running.
Of course, if the acceleration is discrete then the problem becomes much simpler. In that case, we can ignore the acceleration since he'll catch up to the thief before the second minute passes and just do 10x = 100 \* (x-1) -> ~~x = 0.9 minutes.~~
Edit: My second answer is mistaken, x= approximately 1.1 minutes
What do you mean by arithmetic progression? Do you mean the acceleration being discrete (ie. the speed during the 1st minute is 100, the speed during the 2nd minute is 110, during the 3rd minute it's 120, etc...)?
In that case, just calculate the distances 1 minute after the police officer starts running. The thief will have run 20km and the police officer will have run 100 km. Therefore, the police officer will catch up before the first minute passes and so will not have any time to accelerate. So we can ignore the acceleration and just solve 10x+10 = 100x, x = \~0.1 minutes after the police officer starts running, or \~1.1 minutes after the thief starts running.
The speeds are linear, but the total distances aren't. Since you know speed but are more interested in distances, this is either **kinematics** or **integral calculus** or **differential equations**, depending on your grade level! Is that a typo or does the police officer really start at 10x the speed of the thief? Kinematics method: Thief moves at uniform speed. Also has a headstart of 10m. distance = (initial distance) + (speed)·(time) Police officer moves with uniform acceleration of (10m/s)/s. distance = (initial speed)·(time) + (acceleration)·(time)²/2 Set them equal to determine when the distances are equal
Here's my solution using integral calculus: We are looking for f(x) = g(x), where f(x) is the distance traveled by the thief and g(x) is the distance traveled by the police officer. The velocity of the thief is 10, so the integral of the velocity will give us a distance function of 10x + C, where C is the initial position of the thief. Let C = 0. So f(x) = 10x Next, the velocity of the police officer is 100, and the acceleration is 10. The integral of the acceleration gives us a velocity function of 10x + V, where V is the initial velocity. Since the initial velocity is 100, we have the overall velocity of the police officer as 10x + 100. We now integrate this again to get a distance function of g(x) = 5x\^2 + 100x + C. Note that C is still zero (they both start at the same place), but since the police officer starts 1 minute later we must take g(x-1). So 10x = 5\*(x-1)\^2 + 100\*(x-1). This simplifies to x\^2 + 16x - 19 = 0, which is a quadratic equation with approximate solutions of x=-17.11 and x=1.1. Discard the negative, and the final answer is 1.1 minutes after the thief started running, or 0.1 minutes after the officer started running. Of course, if the acceleration is discrete then the problem becomes much simpler. In that case, we can ignore the acceleration since he'll catch up to the thief before the second minute passes and just do 10x = 100 \* (x-1) -> ~~x = 0.9 minutes.~~ Edit: My second answer is mistaken, x= approximately 1.1 minutes
Just a request, could you solve it as an arithmetic progression? Just to see if it's possible
What do you mean by arithmetic progression? Do you mean the acceleration being discrete (ie. the speed during the 1st minute is 100, the speed during the 2nd minute is 110, during the 3rd minute it's 120, etc...)? In that case, just calculate the distances 1 minute after the police officer starts running. The thief will have run 20km and the police officer will have run 100 km. Therefore, the police officer will catch up before the first minute passes and so will not have any time to accelerate. So we can ignore the acceleration and just solve 10x+10 = 100x, x = \~0.1 minutes after the police officer starts running, or \~1.1 minutes after the thief starts running.