contestada

police officer at rest at the side of the highway
notices a speeder moving at 62 km/h along a
straight level road near an elementary school. When
the speeder passes, the officer accelerates at 3.0 m/s2
in pursuit. The speeder does not notice until the
police officer catches up.
Now assume that the police officer accelerates
until the police car is moving 10.0 km/h faster
than the speeder and then moves at a constant
velocity until the police officer catches up. How
long will it take to catch the speeder?

Respuesta :

Let's take the police officers starting position as our origin, that is Sp(t0) = Ss(t0) = 0, and, since we know the police officers starts from a standstill, Vp(t0) = 0 and Vs(t0) = 62km/h

For the first part of the pursuit, during the officers acceleration:

Sp(t) = Sp(t0) + Vp(t0)*t + (3/2)*t^2

Sp(t) = 1.5*t^2

Vp(t) = 3*t

We need to know how much time transpires during this first bit, that is until the officers speed is 72 km/h or 20m/s, and we can find that out by using Vp(t):

Vp(t1) = 20 = 3*t1

t1 = 3.33... s

We now look at how much space each of them covers during this time:

Sp(t1) = 1.5*(t1)^2 = 16.66... m

Ss(t1) = Vs*t1 = (62/3.6)*(3.33...) = 57.41 m

For the second part of the pursuit we will take the officers position as our origin:

Sp(t) = 20*t

Ss(t) = (57.41 - 16.66) + (17.22...)*t

Now we just need to equate Sp(t) to Ss(t):

20*t = (40.75) + (17.22...)*t

t = 40.75/(2.77...) = 14.67 s

But we need to add the time in the first part of the pursuit to the time we just found:

14.67 + 3.33 = 18s

The pursuit lasts 18 seconds.


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