Answer:
(a) 15 m
(b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s
Step-by-step explanation:
We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...
d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².
(a) Then the distance between cars at t=2 is ...
d(2) = 25 - (5/2)·4 = 15 . . . . meters
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(b) The <em>relative velocity</em> at 2.4 seconds is ...
d'(t) = -5t
d'(2.4) = -5(2.4) = -12.0 . . . m/s
The closure speed with the police car is 12 m/s.
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We need to do a little more work to find the ground speed of the trailing car when the cars bump.
The distance between the cars at t=2.4 seconds is ...
d(2.4) = 25 -(5/2)2.4² = 10.6 . . . . meters
At the closure speed of 12 m/s, it will take an additional ...
10.6 m/(12 m/s) = 0.8833... s = 53/60 s
to close the gap. The speed of the trailing car at that point in time will be the original speed less the deceleration for 0.8833 seconds:
(110 km/h)·(1/3.6 (m/s)/(km/h))-(5 m/s²)(53/60 s)
= 26 5/36 m/s = 94.1 km/h
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<em>Comment on following distance</em>
To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.
nor 
are squares. So none of (1), (2), and (4) can be correct. The answer must be (3).