Mike K.

asked • 06/25/20

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A


An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2. The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2. (Hint: The police will not go against the law.)


a. Find the total time required for the police car to overtake the automobile.


b) Find the total distance travelled by the police car while overtaking the automobile.


c) Find the speed of the police car at the time it overtakes the automobile.


d) Find the speed of the automobile at the time it was overtaken by the police car.


Stanton D.

Hi Mike K., problem statement does not include following crucial info: was the police car stationary when the speeding car passed it? If so, and if the police car is travelling exactly along the straight road (so how come the motorist didn't notice the police car?), this is a simple kinematics problem. Motorist travels at constant velocity for 12 s (d=vt); then motorist accelerates at constant rate, (d=d(12)+v(0)(t-12) [i.e. projected motion at constant velocity] - (1/2)a(t-12)^2) [i.e. change due to the acceleration]. Police car is just d=(1/2)at^2. Set the 2 d's equal and solve the quadratic for t. The rest of the stuff is just calculations. -- Cheers, -- Mr. d. P.S. What might be problematic for you, is transforming the word problem correctly into that equation? But take it slowly, and consider how *position* builds on *previous position* , *constant velocity effects*, and *acceleration effects*. Those are the 3 terms of the motorist's equation, with the added twist that they variously occur over the whole time period, or during the time *after* 12 s only!
Report

06/26/20

1 Expert Answer

By:

Sidney P. answered • 06/26/20

Tutor
4.9 (1,532)

Astronomy, Physics, Chemistry, and Math Tutor

See tutors like this
See tutors like this

I tried the quadratic approach outlined in the comment but got a time of 21.1s which is after the auto had stopped, so it wasn't following the Δx = vot = 1/2 at2 anymore! From v = vo + at, I get Δt = 5.13 s for the car to stop after 12 s had gone by. Total distance dc traveled by the car is 15.65*12 + (v2 - vo2)/2a. The second term here is (0 - 15.652)/(-2*3.05) = 40.15, so the police car has to travel 227.95 m total.

For the police car to accelerate from 0 to 11.18 m/s at 1.96 m/s2, 5.70 s are required, and a distance of 31.89 m is covered. The remaining distance to get to the auto is 196.06 m, requiring an additional 17.54 s at the posted speed, so

a) total time for police car is 23.24 s.

b) 227.95 m as above.

c) 11.18 m/s as hinted, since the acceleration phase was only 5.7 s.

d) zero -- the auto had already stopped.

Upvote • 0 Downvote
More
Report

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.