302 DIFFERENTIAL CALCULUS
height reached, (d) the velocity with which
the missile strikes the ground.
[
(a) 100 m/s(b)4s(c) 200 m (d)−100 m/s]- The distancesmetres travelled by a car int
seconds 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) the distance the car travels before it stops.
[(a) 90 km/h (b) 62.5 m] - 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) 4 s (b) 3 rads] - 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 1.5 s (c) the time when the velocity is
zero. ⎡
⎢
⎢
⎣(a) 3 m/s;−1m/s^2(b) 6 m/s;−4m/s^2(c)^34 s⎤⎥
⎥
⎦- The angular displacementθof a rotating disc
is given byθ=6 sin
t
4, wheretis the time in
seconds. Determine (a) the angular velocity
of the disc whentis 1.5 s, (b) the angular
acceleration whentis 5.5 s, and (c) the first
time when the angular velocity is zero.
⎡⎢
⎣(a)ω= 1 .40 rad/s(b)α=− 0 .37 rad/s^2(c)t= 6 .28 s⎤⎥
⎦- x=
20 t^3
3−23 t^2
2+ 6 t+5 represents the dis-
tance,xmetres, moved by a body intseconds.
Determine (a) the velocity and acceleration
at the start, (b) the velocity and acceleration
whent=3 s, (c) the values oftwhen the
body is at rest, (d) the value oftwhen theacceleration is 37 m/s^2 and (e) the distance
travelled in the third second.
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
(a) 6 m/s;−23 m/s^2(b) 117 m/s; 97 m/s^2
(c)^34 sor^25 s
(d) 1^12 s
(e) 75^16 m⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦28.3 Turning points
In Fig. 28.4, the gradient (or rate of change) of the
curve changes from positive betweenOandPto
negative betweenPandQ, and then positive again
betweenQandR. 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 asP
andQare given the general name ofturning points.Figure 28.4It is possible to have a turning point, the gradient
on either side of which is the same. Such a point is
given the special name of apoint of inflexion, and
examples are shown in Fig. 28.5.
Maximum and minimum points and points of
inflexion are given the general term ofstationary
points.
Procedure for finding and distinguishing
between stationary points:(i) Giveny=f(x), determinedy
dx(i.e.f′(x))