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anonymous

  • 4 years ago

A car traveling along the highway passes a point at 20 m/s. After 20 s, a police car passes the same point traveling at 30 m/s. how far down the road, and in what time will the police reach the first car?

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  1. anonymous
    • 4 years ago
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    800m,40secs

  2. anonymous
    • 4 years ago
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    Can you explain how you solved it?

  3. anonymous
    • 4 years ago
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    You could calculate the distance between them at the time when a police car passes that point. The bad guy will be at the distance S_1=v_1*t_1 from the point of interest, and since the police is at that point at that time that means that the distance between them is \[\ S_1=v_1t_1 \] v_1 - velocity of the 1st car t_1 - time passed Lets say you are in the bad guys car, you are moving 20m/s and the police is chasing you with the speed at 30m/s. So, the police car is actually moving 10m/s from your point of view, since you are already moving 20m/s relative to the ground. How much time must pass for police to reach the bad guy? This time will just be the distance between them over their relative velocity, so: \[\ t_2=\frac{S_1}{v_{relative}} \] t_2-time it takes to reach the 1st car v_relative - speed of the police car relative to the 1st car Once you get the time it takes for police to reach the first car, you can easily calculate how far down the road it is. What do you mean by "how far down the road?". Is it from the initial point of interest or from the point where the 1st car was at the time when the police car was at the initial point?

  4. anonymous
    • 4 years ago
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    the point ( Distance and time) at which the police car and the car meet each other.

  5. anonymous
    • 4 years ago
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    Well, anyways, it is easy to calculate it now.

  6. anonymous
    • 4 years ago
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    im still sort of confused. ill try it again.

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