Need help with velocity and accel. problem

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In summary, the police officer will catch up to the speeding car after 20 seconds of pursuit, not 10 seconds as initially calculated. This is because the time should be calculated based on their positions being the same, not their velocities. Using the equation Δx = vt, we can see that the police officer will catch up after traveling a distance of 3t^2, which in this case is 60 meters, in 20 seconds.
  • #1
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Hi there,

I can't seem to understand what I'm doing wrong in the following problem:

"a police officer begins pursuing a speeding car when it passes. If the car moves with a constant velocity of 60.0m/s, and the officer moves with a constant acceleration of 6.0m/s^2. How long will it take the officer to catch up to the car if he begins pursuing from rest?"

I used V(final) = V(initial) + at. Substituting 60.0 for v(final), 0 for v(initial), and 6.0 for a. I solved for t and got 10.0s, but the answer is supposed to be 20.0s. What am I doing wrong?

thanks in advance
 
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  • #2
"Catching up to the car" does not mean that the officer and the speeder have the same speed at the same time; they could be traveling at the same speed, but the officer could be behind (or ahead) of the speeder. You need to solve for the time at which both of them have the same position, not the same speed.
 
  • #3
a police officer begins pursuing a speeding car when it passes. If the car moves with a constant velocity of 60.0m/s, and the officer moves with a constant acceleration of 6.0m/s^2. How long will it take the officer to catch up to the car if he begins pursuing from rest?"

I used V(final) = V(initial) + at. Substituting 60.0 for v(final), 0 for v(initial), and 6.0 for a. I solved for t and got 10.0s, but the answer is supposed to be 20.0s. What am I doing wrong?


You have found the instant of time when the velocities of the two cars will become the same. At that instant the two may not be in the same position.

You have to find the instant when they are in the same POsition, i.e. the instant when their displacements or distance traveled (from the point where the crimnal's car passes the stationary police car), are the same.
 
  • #4
I would suggest drawing a vector diagram. Like u solve other questions in relatve velocity.
I agree with Ambitwistor
 
  • #5
"a police officer begins pursuing a speeding car when it passes. If the car moves with a constant velocity of 60.0m/s, and the officer moves with a constant acceleration of 6.0m/s^2. How long will it take the officer to catch up to the car if he begins pursuing from rest?"


Initial coinciding point is our space origin, the coinciding instant, our time origin.
The path of motion is our X Axis.
X coordinate of criminal's car at any instant t = 60 t meters

X coordinate of police car at any instant t = (1/2)(6)t^2 meters

If the catch up takes place at instant t1;
60(t1) = (1/2)(6)(t1)^2 which has the solutions;
t1 = 0 and t1 = 20.
The first is trivial. The catch up takes place at 20 secs
 
  • #6
Δx = 0.5at^2

a = 6 m/s^2

Δx = 3t^2

Δx = vt

vt = 3t^2

t = v/3 = 60/3 = 20s.
 

FAQ: Need help with velocity and accel. problem

How do I calculate velocity?

To calculate velocity, divide the change in displacement by the change in time. The unit of velocity is meters per second (m/s).

What is the difference between velocity and acceleration?

Velocity is the rate of change of displacement, while acceleration is the rate of change of velocity. In other words, velocity tells us how fast an object is moving, while acceleration tells us how much the velocity is changing.

How do I find the average velocity in a given time interval?

To find the average velocity, divide the total displacement by the total time. This will give you the average velocity over the given time interval.

How do I calculate acceleration?

To calculate acceleration, divide the change in velocity by the change in time. The unit of acceleration is meters per second squared (m/s^2).

What is the meaning of a negative velocity or acceleration?

A negative velocity means that the object is moving in the opposite direction of the chosen positive direction. A negative acceleration means that the object is slowing down in the chosen positive direction.

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