98 Basic Engineering Mathematics
Substituting s = 42 and t = 2 into s = ut +
1
2
at 2 gives
42 = 2u +
1
2
a(2)
2
i.e.
42 = 2u + 2a
(1)
Substituting s = 144 and t = 4 into s = ut +
1
2
at 2
gives
144 = 4u +
1
2
a(4)
2
i.e.
144 = 4u + 8a
(2)
Multiplying equation (1) by 2 gives
84 = 4u + 4a
(3)
Subtracting equation (3) from equation (2) gives
60 = 0 + 4a
and
a =
60
4
= 15
Substituting a = 15 into equation (1) gives
42 = 2u + 2(15)
42 − 30 = 2u
u =
12
2
= 6
Substituting a = 15 and u = 6 in equation (2) gives
RHS = 4(6) + 8(15) = 24 + 120 = 144 = LHS
Hence, the initial velocity u = 6 m/s and the acceleration a = 15 m/s
2 .
Distance travelled after 3 s is given by s = ut +
1
2
at
2
where t = 3, u = 6 and a = 15.
Hence, s = (6)(3) +
1
2
(15)(3) 2 = 18 + 67.5
i.e. distance travelled after 3 s = 85.5 m.
Problem 16. The resistance R of a length of
wire at t ◦ C is given by R = R 0 (1 + αt ), where R 0
is the resistance at 0 ◦ C and α is the temperature
coefficient of resistance in / ◦ C. Find the values of α
and R 0 if R = 30 at 50 ◦ C and R = 35 at 100 ◦ C
Substituting R = 30 and t = 50 into R = R 0 (1 + αt )
gives
30 = R 0 (1 + 50α)
(1)
Substituting R = 35 and t = 100 into R = R 0 (1 + αt )
gives
35 = R 0 (1 + 100α)
(2)
Although these equations may be solved by the conventional substitution method, an easier way is to eliminate
R 0 by division. Thus, dividing equation (1) by equation
(2) gives
30
35
=
R 0 (1 + 50α)
R 0 (1 + 100α)
=
1 + 50α
1 + 100α
Cross-multiplying gives
30(1 + 100α) = 35(1 + 50α)
30 + 3000α = 35 + 1750α
3000α − 1750α = 35 − 30
1250α = 5
i.e.
α =
5
1250
=
1
250
or 0.004
Substituting α =
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, substituting α =
1
250
and R 0 = 25 in equation (2), gives
RHS = 25
1 + (100)
1
250
= 25(1.4) = 35 = LHS
Thus, the solution is α = 0.004/ ◦ C and R 0 = 25 .
Problem 17. The molar heat capacity of a solid
compound is given by the equation c = a + bT ,
where a and b are constants. When c = 52, T = 100
and when c = 172, T = 400. Determine the values
of a and b
Substituting s = 42 and t = 2 into s = ut +
1
2
at 2 gives
42 = 2u +
1
2
a(2)
2
i.e.
42 = 2u + 2a
(1)
Substituting s = 144 and t = 4 into s = ut +
1
2
at 2
gives
144 = 4u +
1
2
a(4)
2
i.e.
144 = 4u + 8a
(2)
Multiplying equation (1) by 2 gives
84 = 4u + 4a
(3)
Subtracting equation (3) from equation (2) gives
60 = 0 + 4a
and
a =
60
4
= 15
Substituting a = 15 into equation (1) gives
42 = 2u + 2(15)
42 − 30 = 2u
u =
12
2
= 6
Substituting a = 15 and u = 6 in equation (2) gives
RHS = 4(6) + 8(15) = 24 + 120 = 144 = LHS
Hence, the initial velocity u = 6 m/s and the acceleration a = 15 m/s
2 .
Distance travelled after 3 s is given by s = ut +
1
2
at
2
where t = 3, u = 6 and a = 15.
Hence, s = (6)(3) +
1
2
(15)(3) 2 = 18 + 67.5
i.e. distance travelled after 3 s = 85.5 m.
Problem 16. The resistance R of a length of
wire at t ◦ C is given by R = R 0 (1 + αt ), where R 0
is the resistance at 0 ◦ C and α is the temperature
coefficient of resistance in / ◦ C. Find the values of α
and R 0 if R = 30 at 50 ◦ C and R = 35 at 100 ◦ C
Substituting R = 30 and t = 50 into R = R 0 (1 + αt )
gives
30 = R 0 (1 + 50α)
(1)
Substituting R = 35 and t = 100 into R = R 0 (1 + αt )
gives
35 = R 0 (1 + 100α)
(2)
Although these equations may be solved by the conventional substitution method, an easier way is to eliminate
R 0 by division. Thus, dividing equation (1) by equation
(2) gives
30
35
=
R 0 (1 + 50α)
R 0 (1 + 100α)
=
1 + 50α
1 + 100α
Cross-multiplying gives
30(1 + 100α) = 35(1 + 50α)
30 + 3000α = 35 + 1750α
3000α − 1750α = 35 − 30
1250α = 5
i.e.
α =
5
1250
=
1
250
or 0.004
Substituting α =
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, substituting α =
1
250
and R 0 = 25 in equation (2), gives
RHS = 25
1 + (100)
1
250
= 25(1.4) = 35 = LHS
Thus, the solution is α = 0.004/ ◦ C and R 0 = 25 .
Problem 17. The molar heat capacity of a solid
compound is given by the equation c = a + bT ,
where a and b are constants. When c = 52, T = 100
and when c = 172, T = 400. Determine the values
of a and b