Example 2-17 depicts the following scenario with the accompanying figures. A spe

Question

Example 2-17 depicts the following scenario with the accompanying figures.
A speeder doing 40.0 mi/h (about 17.9 m/s) in a 25 mi/h zone approaches a parked police car. The instant the speeder passes the police car, the police begin their pursuit. The speeder maintains a constant velocity, and the police car accelerates with a constant acceleration of 3.7 m/s2.

Now we will consider some slightly different related scenarios to Example 2-17.

Part A

Suppose the speeder (red car) is traveling with a constant speed of 22 m/s. If the police car is to start from rest and catch the speeder in 11 s or less, what is the maximum head-start distance the speeder can have? Measure time from the moment the police car starts.   

Part B

Suppose the speeder in part A passes the position of the police car and the police car immediately starts from rest and pursues the speeder with constant acceleration. What acceleration must the police car have if it is to catch the speeder in 8.0 s?

Explanation / Answer

A]

for the police car, u = 0 m/s

and S = ut + (1/2)at2

S = (1/2)(3.7)(11)2 = 223.85 m.

the distance covered by the speeder in the same time is:

S' = 22(11) = 242 m

so, the maximum head start distance which the speeder can have is: S' - S = 18.15 m.

B] for the police, at t=8s,

S = (1/2)a(8)2 = 32a

and in this time, S' = 22 x 8 = 176 m

the distance between them = 18.15 m

so, 18.15 = 176 - 32a

=> a = 4.933 m/s2.

this is the required acceleration for the police car.

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.