Higher Engineering Mathematics

SOME APPLICATIONS OF DIFFERENTIATION 301

G

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 .8 m/s^2

When timet=2s,

velocity,v=(9.8)(2)= 19 .6m/s

and accelerationa= 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 dis-

tance 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=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,

givingt=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 =60 m

Problem 8. The angular displacementθradi-

ans of a flywheel varies with timetseconds

and follows the equationθ= 9 t^2 − 2 t^3. Deter-

mine (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 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 =12 rad/s

Angular accelerationα=

d^2 θ

dt^2

= 18 − 12 trad/s^2

When timet=1s,

α= 18 −12(1)=6 rad/s^2

(b) When the angular acceleration 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 .2 cos 5πt+ 3 .6 sin 5πt. Evaluate the

velocity (in m/s) when timet=30 ms.

Displacementx= 2 .2 cos 5πt+ 3 .6 sin 5πt

Velocityv=

dx

dt

=(2.2)(− 5 π) sin 5πt+(3.6)(5π) cos 5πt

=− 11 πsin 5πt+ 18 πcos 5πtcm/s

When timet=30 ms, velocity

=− 11 πsin

(

5 π·

30

103

)

+ 18 πcos

(

5 π·

30

103

)

=− 11 πsin 0. 4712 + 18 πcos 0. 4712

=− 11 πsin 27◦+ 18 πcos 27◦

=− 15. 69 + 50. 39 = 34 .7 cm/s

=0.347 m/s

Now try the following exercise.

Exercise 123 Further problems on velocity

and acceleration

1. A missile fired from ground level rises
   xmetres vertically upwards intseconds and

x= 100 t−

25

2

t^2. Find (a) the initial velocity

of the missile, (b) the time when the height of

the missile is a maximum, (c) the maximum