Calculating Distance Traveled: Police Chase Scenario

[jrk613]

Homework Statement

A car traveling at a constant speed of 129 km/hr passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets in a chase after the car with a constant acceleration of 2.7 m/s2. How far does the trooper travel before he overtakes the speeding car?

Homework Equations

d= v(initial)t + .5at^2

The Attempt at a Solution

d= 35.83m/s(t) + .5(2.7)(t)^2 + 35.83

I don't understand what needs done, thanks

 

[PhaseShifter]

What is the position of the trooper at time t?

What's the position of his target at time t?

When are they in the same place?

 

[tomcue]

for the help!

I would approach this problem by first identifying the variables and their units. The given information tells us that the initial speed of the car is 129 km/hr, which needs to be converted to m/s (35.83 m/s). The trooper's acceleration is given as 2.7 m/s^2. The time is not given, but we can assume it is the same for both the car and the trooper, as they start their motion at the same time.

Next, I would use the equation d= v(initial)t + .5at^2 to solve for the distance traveled by the trooper. We know the initial velocity (0 m/s) and the acceleration (2.7 m/s^2), so we just need to solve for time. In this case, we can use the formula t= (v(final) - v(initial))/a, where v(final) is the final velocity of the trooper, which we can calculate by adding the trooper's acceleration to the initial speed of the car.

So, t= (35.83 m/s - 0 m/s)/2.7 m/s^2 = 13.3 seconds.

Now, we can plug this value for time into the original equation to solve for distance: d= (35.83 m/s)(13.3 s) + .5(2.7 m/s^2)(13.3 s)^2 = 477.28 meters.

Therefore, the trooper travels 477.28 meters before overtaking the speeding car.