Higher Engineering Mathematics, Sixth Edition

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