Basic Engineering Mathematics, Fifth Edition

98 Basic Engineering Mathematics

Substitutings=42 andt=2intos=ut+

1

2

at^2 gives

42 = 2 u+

1

2

a( 2 )^2

i.e. 42 = 2 u+ 2 a (1)

Substitutings=144 andt=4intos=ut+

1

2

at^2

gives

144 = 4 u+

1

2

a( 4 )^2

i.e. 144 = 4 u+ 8 a (2)

Multiplying equation (1) by 2 gives

84 = 4 u+ 4 a (3)

Subtracting equation (3) from equation (2) gives

60 = 0 + 4 a

and a=^60

4

= 15

Substitutinga=15 into equation (1) gives

42 = 2 u+ 2 ( 15 )

42 − 30 = 2 u

u=

12

2

= 6

Substitutinga=15 andu=6 in equation (2) gives

RHS= 4 ( 6 )+ 8 ( 15 )= 24 + 120 = 144 =LHS

Hence,the initial velocityu=6 m/s and the acceler-

ationa=15m/s^2.

Distance travelled after 3s is given bys=ut+

1

2

at^2

wheret= 3 ,u=6anda=15.

Hence,s=( 6 )( 3 )+

1

2

( 15 )( 3 )^2 = 18 + 67. 5

i.e.distance travelled after 3s= 85 .5m.

Problem 16. The resistanceRof a length of

wire att◦CisgivenbyR=R 0 ( 1 +αt),whereR 0

is the resistance at 0◦Candαis the temperature

coefficient of resistance in /◦C. Find the values ofα

andR 0 ifR= 30 at 50◦CandR= 35 at 100◦C

SubstitutingR=30 andt=50 intoR=R 0 ( 1 +αt)

gives

30 =R 0 ( 1 + 50 α) (1)

SubstitutingR=35 andt=100 intoR=R 0 ( 1 +αt)

gives

35 =R 0 ( 1 + 100 α) (2)

Although these equations may be solved by the conven-

tional substitutionmethod, an easier way is to eliminate

R 0 by division. Thus, dividing equation (1) by equation

(2) gives

30

35

=

R 0 ( 1 + 50 α)

R 0 ( 1 + 100 α)

=

1 + 50 α

1 + 100 α

Cross-multiplying gives

30 ( 1 + 100 α)= 35 ( 1 + 50 α)

30 + 3000 α= 35 + 1750 α

3000 α− 1750 α= 35 − 30

1250 α= 5

i.e. α=^5

1250

=

1

250

or 0. 004

Substitutingα=

1

250

into equation (1) gives

30 =R 0

{

1 +( 50 )

(

1

250

)}

30 =R 0 ( 1. 2 )

R 0 =

30

1. 2

= 25

Checking, substitutingα=

1

250

andR 0 =25 in equa-

tion (2), gives

RHS= 25

{

1 +( 100 )

(

1

250

)}

= 25 ( 1. 4 )= 35 =LHS

Thus, the solution isα= 0. 004 /◦CandR 0 = 25 .

Problem 17. The molar heat capacity of a solid

compound is given by the equationc=a+bT,

whereaandbare constants. Whenc= 52 ,T= 100

and whenc= 172 ,T=400. Determine the values

ofaandb