HELP with Police + Speeding Car :velocity, acceleration, displacement

In summary, the problem involves a police officer trying to catch a car traveling at 85 mi/h on an interstate highway. The squad car's acceleration follows the equation a=a'-bt^2, where a'=2.5m/s^2 and b=0.0028m/s^4. The officer must determine if they can catch the car before it reaches the next exit 5.3 mi away. The attempt to solve the problem involves using the integral of acceleration to find velocity and the integral of velocity to find distance. However, the equation used, x=(a't^2)/2 - (bt^4)/12 + v't +x', results in the wrong answer of 3.8mi. Suggestions
  • #1
jryan422
5
0
1. The problem and given data

You are a police officer and your squad car is at rest on the
shoulder of an interstate highway when you notice a car passing you at
its top speed of 85 mi/h. You jump in your car, start the engine,
and find a break in the traffic, a process which takes 25 s. You
know from the squad car's manual that when it starts from rest
with its accelerator pressed to the floor, the magnitude of its
acceleration is a=a'-bt^2; (where a'=2.5m/s^2 and b=0.0028m/s^4) until a'=bt^2 and then remains zero thereafter.

Can you catch the car before it reaches the next exit 5.3 mi away?2. Any relevant equations

a=a'-bt^2, where a'=2.5m/s^2 and b=0.0028m/s^4

3. The attempt

I'm having trouble using the correct values. I took the integral of acceleration, to obtain velocity, then I took the integral of velocity to obtain distance.

My reasoning is that if I can find the distance, I can find out the exact value before or after 5.3 mi.

My equation ultimately is: x=(a't^2)/2 - (bt^4)/12 + v't +x'

where a', t and b are given. I used 85 mi/hr for v' and x' =0. I get the wrong answer.
The right answer is 3.8mi.

Any advice/suggestions will be greatly appreciated for this struggling physics student :frown:
 
Last edited:
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  • #2
i have a question similar to this on my web homework that's due in 28 minutes
 
  • #3
ok, well is there a different way you set up the problem?
 
  • #4
sorry, but anyone?
 

What is the difference between velocity and acceleration?

Velocity is the rate of change of an object's displacement over time, while acceleration is the rate of change of an object's velocity over time. In simpler terms, velocity tells us how fast and in what direction an object is moving, while acceleration tells us how quickly the velocity is changing.

How can I calculate the velocity of a speeding car using displacement and time?

The velocity of a speeding car can be calculated by dividing the displacement (distance traveled) by the time it took to travel that distance. So if a car traveled 100 meters in 10 seconds, its velocity would be 100 meters / 10 seconds = 10 meters per second.

Why is it important for police to measure the acceleration of a speeding car?

Measuring the acceleration of a speeding car can help police determine how quickly the car was traveling and how long it took to reach that speed. This information can be crucial in determining the severity of the offense and any potential dangers to other drivers on the road.

What units are typically used to measure velocity and acceleration?

Velocity is typically measured in meters per second (m/s) or kilometers per hour (km/h), while acceleration is measured in meters per second squared (m/s²). However, other units such as miles per hour (mph) and feet per second squared (ft/s²) may also be used depending on the specific situation.

Is it possible for a car to have a high velocity but low acceleration?

Yes, it is possible for a car to have a high velocity but low acceleration. This can occur when a car is traveling at a constant speed, meaning its velocity is not changing, so the acceleration would be 0. Alternatively, a car can have a high velocity and low acceleration if it is decelerating, meaning its velocity is decreasing over time.

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