Problem 2-101 - Linear Kinematics - Part 6 - (b)

A car \((c)\) with one headlight burned out is traveling at a constant speed of \(18 \;m/s\) and passes a stopped police car \((p)\). The car is pursued immediately by the police cruiser, which has a constant acceleration of magnitude \(2.2 \;m/s^2.\)

(a) How far does the police cruiser travel before catching the other car?
(b) At what time will this occur? (Hint: Graphing may help to visualize this problem.)

Accumulated Solution

Displacement of the car
\(x_c = x_{0c} + v_{0c^t}\)

Displacement of the cruiser
\(x_p = x_{0p} + v_{0p} + ½ \;at^2\)

Correct!
Set \(x_p = x_c = x\) and eliminate \(t\)
 
From the first equation \(t = x/v_c\)
Substitute this into the second equation
 
\(x = ½ \;a(x/v_c)^2\)
 
Therefore
\(x = 2v_{c^2}/a = 1(18)^2/(2.2) = 2.9 \times10^2\; m\)
 
This is the answer to part (a)

To find the time elapsed until they meet we should:

(A)  Insert \(x = 290\; m\) in the first equation and solve for \(t.\)

(B)  Insert \(x = 290\; m\) in the second equation and solve for \(t.\)

(C)  Equate the two \((x_c\; \text{and}\;x_p)\) and solve for \(t.\)