302 Higher Engineering Mathematics
Problem 6. Supplies are dropped from a
helicoptor and the distance fallen in a time
t seconds is given by x =
1
2 gt 2 , where g = 9.8 m/s 2 .
Determine the velocity and acceleration of the
supplies after it has fallen for 2 seconds.
Distance
x =
1
2
gt 2 =
1
2
(9.8)t 2 = 4.9t 2 m
Velocity
v =
dv
dt
= 9.8t m/s
and acceleration a =
d 2 x
dt 2 = 9.8 m/s 2
When time t = 2 s,
velocity, v = (9.8)(2) = 19.6 m/s
and acceleration a = 9.8m/s
2
(which is acceleration due to gravity).
Problem 7. The distance x metres travelled by a
vehicle in time t seconds after the brakes are
applied is given by x = 20t −
5
3 t 2 . Determine (a) the
speed of the vehicle (in km/h) at the instant the
brakes are applied, and (b) the distance the car
travels before it stops.
(a) Distance, x = 20t −
5
3 t 2 .
Hence velocity v =
dx
dt
= 20 −
10
3
t .
At the instant the brakes are applied, time = 0.
Hence velocity, v = 20 m/s
=
20 × 60 × 60
1000
km/h
= 72 km/h
(Note: changing from m/s to km/h merely involves
multiplying by 3.6.)
(b) When the car finally stops, the velocity is zero, i.e.
v = 20 −
10
3
t = 0, from which, 20 =
10
3
t , giving
t = 6 s.
Hence the distance travelled before the car stops
is given by:
x = 20t −
5
3 t
2
= 20(6) −
5
3 (6)
2
= 120 − 60 = 60 m
Problem 8. The angular displacement θ radians
of a flywheel varies with time t seconds and follows
the equation θ = 9t 2 − 2t 3 . Determine (a) the
angular velocity and acceleration of the flywheel
when time, t = 1 s, and (b) the time when the
angular acceleration is zero.
(a) Angular displacement θ = 9t
2
− 2t
3 rad
Angular velocity ω =
dθ
dt
= 18t − 6t 2 rad/s
When time t = 1 s,
ω = 18(1) − 6(1)
2
= 12 rad/s
Angular acceleration α =
d 2 θ
dt 2 = 18 − 12t rad/s 2
When time t = 1 s,
α = 18 − 12(1) = 6 rad/s
2
(b) When the angular acceleration is zero,
18 − 12t = 0, from which, 18 =12t , giving time,
t = 1.5 s.
Problem 9. The displacement x cm of the slide
valve of an engine is given by
x = 2.2 cos 5πt + 3.6 sin5πt . Evaluate the
velocity (in m/s) when time t = 30 ms.
Displacement x = 2.2 cos 5πt + 3.6 sin5πt
Velocity v =
dx
dt
= (2.2)(−5π)sin 5πt + (3.6)(5π)cos 5πt
= −11π sin 5πt + 18π cos 5πt cm/s
When time t = 30 ms, velocity
= −11π sin
5π ·
30
10 3
+ 18π cos
5π ·
30
10 3
= −11π sin 0.4712 + 18π cos 0.4712
= −11π sin 27
◦
+ 18π cos 27
◦
= −15.69 + 50.39 = 34.7 cm/s
= 0.347 m/s
Problem 6. Supplies are dropped from a
helicoptor and the distance fallen in a time
t seconds is given by x =
1
2 gt 2 , where g = 9.8 m/s 2 .
Determine the velocity and acceleration of the
supplies after it has fallen for 2 seconds.
Distance
x =
1
2
gt 2 =
1
2
(9.8)t 2 = 4.9t 2 m
Velocity
v =
dv
dt
= 9.8t m/s
and acceleration a =
d 2 x
dt 2 = 9.8 m/s 2
When time t = 2 s,
velocity, v = (9.8)(2) = 19.6 m/s
and acceleration a = 9.8m/s
2
(which is acceleration due to gravity).
Problem 7. The distance x metres travelled by a
vehicle in time t seconds after the brakes are
applied is given by x = 20t −
5
3 t 2 . Determine (a) the
speed of the vehicle (in km/h) at the instant the
brakes are applied, and (b) the distance the car
travels before it stops.
(a) Distance, x = 20t −
5
3 t 2 .
Hence velocity v =
dx
dt
= 20 −
10
3
t .
At the instant the brakes are applied, time = 0.
Hence velocity, v = 20 m/s
=
20 × 60 × 60
1000
km/h
= 72 km/h
(Note: changing from m/s to km/h merely involves
multiplying by 3.6.)
(b) When the car finally stops, the velocity is zero, i.e.
v = 20 −
10
3
t = 0, from which, 20 =
10
3
t , giving
t = 6 s.
Hence the distance travelled before the car stops
is given by:
x = 20t −
5
3 t
2
= 20(6) −
5
3 (6)
2
= 120 − 60 = 60 m
Problem 8. The angular displacement θ radians
of a flywheel varies with time t seconds and follows
the equation θ = 9t 2 − 2t 3 . Determine (a) the
angular velocity and acceleration of the flywheel
when time, t = 1 s, and (b) the time when the
angular acceleration is zero.
(a) Angular displacement θ = 9t
2
− 2t
3 rad
Angular velocity ω =
dθ
dt
= 18t − 6t 2 rad/s
When time t = 1 s,
ω = 18(1) − 6(1)
2
= 12 rad/s
Angular acceleration α =
d 2 θ
dt 2 = 18 − 12t rad/s 2
When time t = 1 s,
α = 18 − 12(1) = 6 rad/s
2
(b) When the angular acceleration is zero,
18 − 12t = 0, from which, 18 =12t , giving time,
t = 1.5 s.
Problem 9. The displacement x cm of the slide
valve of an engine is given by
x = 2.2 cos 5πt + 3.6 sin5πt . Evaluate the
velocity (in m/s) when time t = 30 ms.
Displacement x = 2.2 cos 5πt + 3.6 sin5πt
Velocity v =
dx
dt
= (2.2)(−5π)sin 5πt + (3.6)(5π)cos 5πt
= −11π sin 5πt + 18π cos 5πt cm/s
When time t = 30 ms, velocity
= −11π sin
5π ·
30
10 3
+ 18π cos
5π ·
30
10 3
= −11π sin 0.4712 + 18π cos 0.4712
= −11π sin 27
◦
+ 18π cos 27
◦
= −15.69 + 50.39 = 34.7 cm/s
= 0.347 m/s