Higher Engineering Mathematics, Sixth Edition

Some applications of differentiation 303

Now try the following exercise

Exercise 121 Further problemson 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

height reached, (d) the velocity with which the

missile strikes the ground.

[

(a)100m/s (b)4s

(c)200m (d)−100m/s

]

2. The distance s metres travelled by a car
   intseconds after the brakes are applied is
   given bys= 25 t− 2. 5 t^2. Find (a) the speed
   of the car (in km/h) when the brakes are
   applied,(b) thedistancethecar travels beforeit
   stops. [(a) 90km/h (b) 62.5m]


3. The equationθ= 10 π+ 24 t− 3 t^2 gives the
   angleθ, in radians, through which a wheel
   turns intseconds. Determine (a) the time
   the wheel takes to come to rest, (b) the
   angle turned through in the last second of
   movement. [(a) 4s (b) 3rads]


4. At any timetseconds the distancexmetres
   of a particle moving in a straight line from
   a fixed point is given byx= 4 t+ln( 1 −t).
   Determine (a) the initial velocity and
   acceleration (b) the velocity and acceleration
   after 1.5s (c) the time when the velocity is
   zero.
   ⎡
   ⎢
   ⎢
   ⎣

(a)3m/s;−1m/s^2

(b)6m/s;−4m/s^2

(c)^34 s

⎤

⎥

⎥

⎦

5. Theangulardisplacementθofarotatingdiscis
   given byθ=6sin

t

4

,wheretis thetime in sec-

onds. Determine (a) the angular velocityof the

discwhentis1.5s,(b)theangularacceleration

whentis 5.5s, and (c) the first time when the

angular velocity is zero.

⎡

⎢

⎣

(a)ω= 1 .40rad/s

(b)α=− 0 .37rad/s^2

(c)t= 6 .28s

⎤

⎥

⎦

6. x=

20 t^3

3

−

23 t^2

2

+ 6 t+5 represents the dis-

tance,xmetres, moved by a bodyintseconds.

Determine (a) the velocity and acceleration

at the start, (b) the velocity and acceleration

whent=3s, (c) the values oftwhen the

body is at rest, (d) the value oftwhen the

acceleration is 37m/s^2 and (e) the distance

travelled in the third second.

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(a)6m/s;−23m/s^2

(b)117m/s;97m/s^2

(c)^34 sor^25 s

(d) 112 s

(e) 7516 m

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

28.3 Turning points

In Fig. 28.4, the gradient (or rate of change) of the

curve changes from positive between O and P to

negative betweenP andQ, and then positive again

betweenQ andR. At pointP, the gradient is zero

and, asxincreases, the gradient of the curve changes

from positive just beforePto negative just after. Such

a point is called amaximum pointand appears as the

‘crest of a wave’. At pointQ, the gradient is also zero

and, asxincreases, the gradient of the curve changes

from negative just beforeQto positive just after. Such

a point is called aminimum point, and appears as the

‘bottom of a valley’. Points such asPandQare given

the general name ofturning points.

O Q

y

P

R

x

Positive

gradient

Positive

gradient

Negative

gradient

Figure 28.4