Integration and Kinematics.

[211 marks]

A body moves in a straight line such that its velocity, v ms−1 , after t seconds is
t
given by v = 2 sin ( 10 + π5 ) csc( 30t + π4 ) for 0 ⩽ t ⩽ 60.
The following diagram shows the graph of v against t. Point A is a local maximum
and point B is a local minimum.

1a. Determine the coordinates of point A and the coordinates of point B. [4 marks]

1b. Hence, write down the maximum speed of the body. [1 mark]

The body first comes to rest at time t = t1 . Find

1c. the value of t1 . [2 marks]

1d. the distance travelled between t = 0 and t = t1 . [2 marks]

1e. the acceleration when t = t1 . [2 marks]

1f. Find the distance travelled in the first 30 seconds. [3 marks]

2a. Show that (sin x + cos x)2 = 1 + sin 2x. [2 marks]

2b. Show that sec 2x + tan 2x = cos x+sin x . [4 marks]

cos x−sin x

2c. π
[9 marks]
Hence or otherwise find ∫0 6
(sec 2x + tan 2x) dx in the form
ln (a + √b) where a, b ∈ Z.

3. Using the substitution cos 3x dx [5 marks]

u = sin x, find ∫ .
√sin x

The function f is defined by f (x) = sec x + 2, 0 ⩽ x < π

2.

4. Use integration by parts to find ∫ (ln x)2 dx. [6 marks]

5. A particle moves along a horizontal line such that at time t seconds, t ≥ [6 marks]
0, its acceleration a is given by a = 2t − 1. When t = 6 , its displacement
s from a fixed origin O is 18.25 m. When t = 15, its displacement from O is
922.75 m. Find an expression for s in terms of t.

6a. Use integration by parts to show that [5 marks]

2ex ex
∫ ex cos 2xdx = 5
sin 2x + 5
cos 2x + c, c ∈ R.

6b. Hence, show that ∫ ex cos 2 xdx = ex sin 2x + ex cos 2x + ex + c, c ∈ R. [3 marks]

5 10 2

2
 The function f is defined by f (x) = ex cos 2 x, where 0 ≤ x ≤ 5. The curve
y = f (x) is shown on the following graph which has local maximum points at A
and C and touches the x-axis at B and D.

6c. Find the x-coordinates of A and of C , giving your answers in the form [6 marks]
a + arctan b, where a, b ∈ R.

6d. Find the area enclosed by the curve and the x-axis between B and D, as [5 marks]
shaded on the diagram.

Consider the functions f, g, defined for x ∈ R, given by f (x) = e−x sin x and
g (x) = e−x cos x.

7a. Find f ′ (x). [2 marks]

7b. Find g ′ (x). [1 mark]

7c. π [4 marks]
∫
Hence, or otherwise, find 0 e−x sin x dx.

1
8a. Use the substitution u = x 2 to find ∫ dx . [4 marks]
3 1
x 2 +x 2

8b. 9 [3 marks]
∫
Hence find the value of 12 1 dx
3 1 , expressing your answer in the form
x 2 +x 2
arctan q, where q ∈ Q.

−1
 A point P moves in a straight line with velocity v ms−1 given by
v (t) = e−t − 8t2 e−2t at time t seconds, where t ≥ 0.

9a. Determine the first time t1 at which P has zero velocity. [2 marks]

9b. Find an expression for the acceleration of P at time t. [2 marks]

9c. Find the value of the acceleration of P at time t1. [1 mark]

10. A particle moves in a straight line such that at time t seconds (t ⩾ 0), its [5 marks]
velocity v, in ms−1 , is given by v = 10te−2t . Find the exact distance travelled by
the particle in the first half-second.

11. By using the substitution x2 = 2 sec θ, show that [7 marks]

∫ dx = 14 arccos( x22 ) + c.
x√x4−4

12. Find ∫ arcsin x dx [5 marks]

A particle moves along a straight line. Its displacement, s metres, at time t

seconds is given by s = t + cos 2t, t ⩾ 0. The first two times when the particle is
at rest are denoted by t1 and t2 , where t1 < t2 .

13a. Find t1 and t2 . [5 marks]

13b. Find the displacement of the particle when t = t1 [2 marks]

14a. 1
π
4 [4 marks]
∫ ∫
1
Using the substitution x = tan θ show that 0 dx = 0 cos2 θdθ.
(x +1)2
2

14b. 1 [3 marks]
∫
1
Hence find the value of 0 dx.
(x2+1)2
 Xavier, the parachutist, jumps out of a plane at a height of h metres above the
ground. After free falling for 10 seconds his parachute opens. His velocity, v ms−1 ,
t seconds after jumping from the plane, can be modelled by the function
⎧ 9.8t, 0 ⩽ t ⩽ 10
v(t) = ⎨
⎩ √1+( −10)2 ,
98
t > 10
t

15a. Find his velocity when t = 15. [2 marks]

15b. Calculate the vertical distance Xavier travelled in the first 10 seconds. [2 marks]

His velocity when he reaches the ground is 2.8ms−1 .

15c. Determine the value of h . [5 marks]

Let y = ex sin x.

16a. Find an expression for dy . [2 marks]

dx

16b. d 2y [2 marks]
Show that
d x2
= 2ex cos x.

Consider the function f defined by f(x) = ex sin x, 0 ⩽ x ⩽ π.

16c. Show that the function f has a local maximum value when 3π [2 marks]
x= 4
.

16d. Find the x-coordinate of the point of inflexion of the graph of f . [2 marks]

16e. Sketch the graph of f , clearly indicating the position of the local [3 marks]
maximum point, the point of inflexion and the axes intercepts.

16f. Find the area of the region enclosed by the graph of f and the x-axis. [6 marks]
∣∣ d 2y ∣∣
∣ d x2 ∣
The curvature at any point (x, y) on a graph is defined as κ = 3 .
2
(1+( d x ) )
dy 2
 ∣∣ d 2y ∣∣
∣ d x2 ∣
The curvature at any point (x, y) on a graph is defined as κ = 3 .
2
(1+( d x ) )
dy 2

16g. Find the value of the curvature of the graph of f at the local maximum [3 marks]
point.

16h. Find the value κ for x = π2 and comment on its meaning with respect to[2 marks]
the shape of the graph.

(ln x)2
The following diagram shows the graph of y= x
,x > 0.

17a. Given that the curve passes through the point (a, 0), state the value of [1 mark]
a.

The region R is enclosed by the curve, the x-axis and the line x = e.

17b. Use the substitution u = ln x to find the area of the region R. [5 marks]

e (ln x)n
Let In = ∫1 x2 dx, n ∈ N.

17c. (i) Find the value of I0 . [7 marks]

1
(ii) Prove that In = e + nIn−1 , n ∈ Z+ .
(iii) Hence find the value of I1 .
17d. Find the volume of the solid formed when the region R is rotated [5 marks]
through 2π about the x-axis.

The displacement, s, in metres, of a particle t seconds after it passes through the

origin is given by the expression s = ln(2 − e−t ), t ⩾ 0.

18a. Find an expression for the velocity, v, of the particle at time t. [2 marks]

18b. Find an expression for the acceleration, a, of the particle at time t. [2 marks]

18c. Find the acceleration of the particle at time t = 0. [1 mark]

A particle moves such that its velocity v ms−1 is related to its displacement s m ,
by the equation v(s) = arctan(sin s), 0 ⩽ s ⩽ 1. The particle’s acceleration is
a ms−2 .

19a. Find the particle’s acceleration in terms of s. [4 marks]

19b. Using an appropriate sketch graph, find the particle’s displacement [2 marks]
when its acceleration is 0.25 ms−2 .

Richard, a marine soldier, steps out of a stationary helicopter, 1000 m above the
ground, at time t = 0. Let his height, in metres, above the ground be given by
s(t). For the first 10 seconds his velocity, v(t)ms−1 , is given by v(t) = −10t.

20a. (i) Find his acceleration a(t) for t < 10. [6 marks]
(ii) Calculate v(10).
(iii) Show that s(10) = 500.

20b. At t = 10 his parachute opens and his acceleration a(t) is subsequently [1 mark]
given by a(t) = −10 − 5v, t ≥ 10.
1
Given that ddt = dv , write down ddt in terms of v.
v v
dt

( ) = −10 − 5
20c. You are told that Richard’s acceleration, a(t) = −10 − 5v, is always [5 marks]
positive, for t ≥ 10.

Hence show that t = 10 + 15 ln( −2−

98
v
).

20d. You are told that Richard’s acceleration, a(t) = −10 − 5v, is always [2 marks]
positive, for t ≥ 10.
Hence find an expression for the velocity, v, for t ≥ 10.

20e. You are told that Richard’s acceleration, a(t) = −10 − 5v, is always [5 marks]
positive, for t ≥ 10.
Find an expression for his height, s, above the ground for t ≥ 10.

20f. You are told that Richard’s acceleration, a(t) = −10 − 5v, is always [2 marks]
positive, for t ≥ 10.
Find the value of t when Richard lands on the ground.

A particle moves in a straight line, its velocity vms−1 at time t seconds is given by
v = 9t − 3t2 , 0 ≤ t ≤ 5.
At time t = 0, the displacement s of the particle from an origin O is 3 m.

21a. Find the displacement of the particle when t = 4. [3 marks]

21b. Sketch a displacement/time graph for the particle, 0 ≤ t ≤ 5, showing [5 marks]

clearly where the curve meets the axes and the coordinates of the points where
the displacement takes greatest and least values.

21c. For 2πt [3 marks]

t > 5, the displacement of the particle is given by s = a + b cos 5
such that s is continuous for all t ≥ 0.
Given further that s = 16.5 when t = 7.5 , find the values of a and b .

21d. For t > 5, the displacement of the particle is given by s = a + b cos 25πt [4 marks]
such that s is continuous for all t ≥ 0.
Find the times t1 and t2 (0 < t1 < t2 < 8) when the particle returns to its starting
point.
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