302 Higher Engineering Mathematics
Problem 6. Supplies are dropped from a
helicoptor and the distance fallen in a time
tseconds is given byx=^12 gt^2 ,whereg= 9 .8m/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. 9 t^2 m
Velocity v=
dv
dt
= 9. 8 tm/s
and acceleration a=
d^2 x
dt^2
= 9 .8m/s^2
When timet=2s,
velocity,v=( 9. 8 )( 2 )= 19 .6m/s
andaccelerationa= 9 .8m/s^2
(which is acceleration due to gravity).
Problem 7. The distancexmetres travelled by a
vehicle in timetseconds after the brakes are
applied is given byx= 20 t−^53 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= 20 t−^53 t^2.
Hence velocityv=
dx
dt
= 20 −
10
3
t.
At the instant the brakes are applied, time=0.
Hencevelocity,v=20m/s
=
20 × 60 × 60
1000
km/h
=72km/h
(Note: changing from m/s to km/hmerely 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=6s.
Hence the distance travelled before the car stops
is given by:
x= 20 t−^53 t^2 = 20 ( 6 )−^53 ( 6 )^2
= 120 − 60 =60m
Problem 8. The angular displacementθradians
of a flywheel varies with timetseconds and follows
the equationθ= 9 t^2 − 2 t^3. Determine (a) the
angular velocity andacceleration of the flywheel
when time,t=1s, and (b) the time when the
angularacceleration is zero.
(a) Angular displacementθ= 9 t^2 − 2 t^3 rad
Angular velocityω=
dθ
dt
= 18 t− 6 t^2 rad/s
When timet=1s,
ω= 18 ( 1 )− 6 ( 1 )^2 =12rad/s
Angularaccelerationα=
d^2 θ
dt^2
= 18 − 12 trad/s^2
When timet=1s,
α= 18 − 12 ( 1 )=6rad/s^2
(b) When the angularacceleration is zero,
18 − 12 t=0, from which, 18= 12 t, giving time,
t= 1 .5s.
Problem 9. The displacementxcm of the slide
valve of an engine is given by
x= 2 .2cos5πt+ 3 .6sin5πt.Evaluatethe
velocity (in m/s) when timet=30ms.
Displacementx= 2 .2cos5πt+ 3 .6sin5πt
Velocityv=
dx
dt
=( 2. 2 )(− 5 π)sin5πt+( 3. 6 )( 5 π)cos 5πt
=− 11 πsin5πt+ 18 πcos5πtcm/s
When timet=30ms, velocity
=− 11 πsin
(
5 π·
30
103
)
+ 18 πcos
(
5 π·
30
103
)
=− 11 πsin0. 4712 + 18 πcos0. 4712
=− 11 πsin27◦+ 18 πcos27◦
=− 15. 69 + 50. 39 = 34 .7cm/s
=0.347m/s