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