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Exam Room: P101_Index: Student ID Number: Signature:
2nd
Method:
a) Note that the total time of the motion is t1+t2! The acceleration of the police car is
211
2
2
1
2
1
m/s -1a
0m/s 3
tttt
ttaa
We are given that )()( and )()( 21212121 ttvttvttxttx carpolicecarpolice
When t = t1 the position of the cars are111
2
1
2
111 24)(,2
3
2
1)( ttvtxttatx carcarpolice
When t = t1 the velocity of the police car is 1111 3)( ttatvpolice
When t = t1+t2 the position of the cars are
)(24)(
2
13
2
3
2
1)()()(
2121
2
221
2
1
2
2221121
ttttx
tttttattvtxttx
car
policepolicepolice
When t = t1+t2 the velocity of the police car is .243)()( 2122121 tttatvttv policepolice
Put 24 = 3t1 + t2 in the equation for xcar and use xpolice = xcar at t = t1 + t2.
))(3(2
13
2
32121
2
221
2
1 tttttttt
This equation gives
0))(3())(()(2)(2223 21212121211
2
2
2
121
2
1
2
221
2
1 tttttttttttttttttttt .
Now, the last equation has the mathematical solutions 3t1 = -t2 and t1 = t2. Since time is not negative, the only
physical solution is t1 = t2. Then, using 3t1 + t2 = 24 we get t1 = t2 = 12 s.
Then, xpolice = xcar = 24 (t1 + t2) = 576 m.
b) m/s. 363)( 1111 ttatvpolice m/s. 24carpoliceav vv