Sec 4 A-Math Kinematics worksheet with answers

A free Sec 4 A-Math worksheet on kinematics: 5 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Kinematics notes. Or practise it online with step-by-step help when you get one wrong.

Try a question: Kinematics

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

MugCornerT-Leng Tuition · mugcornersg.com

Sec 4 A-Math worksheet: Kinematics

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

(a)A particle moves in a straight line so that,

t

seconds after passing through a fixed point

O

, its velocity,

v

m/s, is given by

v = 9t^{2} - 8t

. Find its displacement from

O

when

t = 3

Answer (a)[3]

(b)A particle moves in a straight line so that,

t

seconds after passing through a fixed point

O

, its acceleration,

a

m/s², is given by

a = 4t + 3

. Its initial velocity is

-3

m/s. Find its velocity when

t = 2

Answer (b)[3]

(a)A particle moves in a straight line so that,

t

seconds after passing through a fixed point

O

, its velocity,

v

m/s, is given by

v = 3t^{2} - 42t + 120

. Find the total distance travelled by the particle in the first

9

seconds.

Answer (a)[3]

(b)A particle moves in a straight line so that,

t

seconds after passing through a fixed point

O

, its velocity,

v

m/s, is given by

v = t^{2} - 8t + 12

. Find the total distance travelled by the particle in the first

5

seconds.

Answer (b)[3]

(a)A particle moves in a straight line so that,

t

seconds after passing through a fixed point

O

, its displacement from

O

s

m, is given by

s = t^{3} - 12t^{2} + 36t

. Find its acceleration when

t = 2

Answer (a)[3]

(b)A particle moves in a straight line so that,

t

seconds after passing through a fixed point

O

, its displacement from

O

s

m, is given by

s = 2t^{3} - 24t^{2} + 42t

. Find the values of

t

at which the particle is instantaneously at rest.

Answer (b)[3]

Section B

A particle moves in a straight line so that,

t

seconds after passing through a fixed point

O

, its velocity,

v

m/s, is given by

v = 12e^{-\frac{t}{2}} - 2

(a)Find the initial velocity of the particle.

Answer (a)m/s[1]

(b)Find the initial acceleration of the particle.

Answer (b)m/s²[2]

(c)Find the exact value of

t

at which the particle is instantaneously at rest.

Answer (c)[2]

(d)Find the displacement of the particle from

O

when it is instantaneously at rest, correct to 3 significant figures.

Answer (d)m[2]

(e)Find the total distance travelled by the particle in the first

4

seconds, correct to 3 significant figures.

Answer (e)m[3]

A particle

P

starts from a fixed point

O

with velocity

-12

m/s and moves in a straight line. Its acceleration,

a

m/s²,

t

seconds after leaving

O

, is given by

a = 8 - 2t

, until it reaches its greatest velocity at

t = T

. After that it decelerates uniformly at

1.5

m/s² until it comes to rest.

(a)Find an expression for the velocity of

P

in terms of

t

, for

0 \leq t \leq T

Answer (a)[2]

(b)Find the time at which

P

is first at instantaneous rest.

Answer (b)s[2]

(c)Find the value of

T

Answer (c)s[1]

(d)Find the greatest velocity of

P

Answer (d)m/s[1]

(e)Find the time at which

P

comes to rest.

Answer (e)s[2]

(f)Find the total distance travelled by

P

Answer (f)m[4]

END OF PAPER

Answers and worked solutions

Question 1

(a)Answer:

= 45

m[3]

s = \dfrac{9t^{3}}{3} - \dfrac{8t^{2}}{2} + c

= 3t^{3} - 4t^{2} + c

0 = 0 + c

c = 0

s = 3(3)^{3} - 4(3)^{2}

= 45

(b)Answer:

= 11

m/s[3]

v = \dfrac{4t^{2}}{2} + 3t + c

= 2t^{2} + 3t + c

-3 = 0 + c

c = -3

v = 2(2)^{2} + 3(2) - 3

= 11

m/s

Question 2

(a)Answer:

= 308

m[3]

3t^{2} - 42t + 120 = 0

3(t - 4)(t - 10) = 0

t = 4

t = 10

It turns back at

t = 4

s = \dfrac{3t^{3}}{3} - \dfrac{42t^{2}}{2} + 120t

= t^{3} - 21t^{2} + 120t

When

t = 4

s = 208

When

t = 9

s = 108

\text{Distance} = \left|208 - 0\right| + \left|108 - 208\right|

\text{Distance} = 208 + 100

= 308

(b)Answer:

= \dfrac{59}{3}

m[3]

t^{2} - 8t + 12 = 0

(t - 2)(t - 6) = 0

t = 2

t = 6

It turns back at

t = 2

s = \dfrac{t^{3}}{3} - \dfrac{8t^{2}}{2} + 12t

= \dfrac{1}{3}t^{3} - 4t^{2} + 12t

When

t = 2

s = \dfrac{32}{3}

When

t = 5

s = \dfrac{5}{3}

\text{Distance} = \left|\dfrac{32}{3} - 0\right| + \left|\dfrac{5}{3} - \dfrac{32}{3}\right|

\text{Distance} = \dfrac{32}{3} + 9

= \dfrac{59}{3}

Question 3

(a)Answer:

= -12

m/s²[3]

v = \dfrac{ds}{dt}

v = 3t^{2} - 12(2)t + 36(1)

= 3t^{2} - 24t + 36

a = \dfrac{dv}{dt}

a = 3(2)t - 24(1)

= 6t - 24

a = 6(2) - 24

= -12

m/s²

(b)Answer:

t = 1

t = 7

[3]

v = \dfrac{ds}{dt}

v = 2(3)t^{2} - 24(2)t + 42(1)

= 6t^{2} - 48t + 42

At rest:

v = 0

6t^{2} - 48t + 42 = 0

t^{2} - 8t + 7 = 0

(t - 1)(t - 7) = 0

t = 1

t = 7

Question 4

(a)Answer:

10

m/s[1]

When

t = 0

v = 12e^{0} - 2

= 10

m/s

(b)Answer:

-6

m/s²[2]

a = \dfrac{dv}{dt}

= -6e^{-\frac{t}{2}}

When

t = 0

a = -6

m/s²

(c)Answer:

t = 2\ln 6

[2]

12e^{-\frac{t}{2}} - 2 = 0

e^{-\frac{t}{2}} = \dfrac{1}{6}

-\frac{t}{2} = -\ln 6

t = 2\ln 6

(d)Answer:

12.8

m[2]

s = \int \left(12e^{-\frac{t}{2}} - 2\right) dt

= -24e^{-\frac{t}{2}} - 2t + c

s = 0

when

t = 0

, so

c = 24

s = 24 - 24e^{-\frac{t}{2}} - 2t

When

t = 2\ln 6

s = 24 - 4 - 2\left(2\ln 6\right)

= 12.8

m (3 s.f.)

(e)Answer:

12.9

m[3]

The particle is at rest at

t = 2\ln 6 \approx 3.584

, and moves back towards

O

after that.When

t = 4

s = 24 - 24e^{-2} - 8

s = 12.752

Distance

= 12.833 + (12.833 - 12.752)

= 12.9

m (3 s.f.)

Question 5

(a)Answer:

v = -12 + 8t - t^{2}

[2]

v = \int (8 - 2t) \, dt

= 8t - t^{2} + c_{1}

v = -12

when

t = 0

, so

c_{1} = -12

v = -12 + 8t - t^{2}

(b)Answer:

t = 2

s[2]

-12 + 8t - t^{2} = 0

t^{2} - 8t + 12 = 0

(t - 2)(t - 6) = 0

t = 2

t = 6

; the first is

t = 2

(and

t = 6

is after

T

)

(c)Answer:

T = 4

s[1]

The velocity is greatest when

a = 0

8 - 2T = 0

T = 4

(d)Answer:

4

m/s[1]

v = -12 + 8(4) - (4)^{2}

= 4

m/s

(e)Answer:

t = \dfrac{20}{3}

s[2]

After

t = 4

v = 4 - 1.5(t - 4)

v = 0

when

t - 4 = \dfrac{4}{1.5}

t - 4 = \dfrac{8}{3}

t = \dfrac{20}{3}

(f)Answer:

\dfrac{64}{3}

m[4]

s = \int (-12 + 8t - t^{2}) \, dt

\phantom{s} = -12t + 4t^{2} - \dfrac{t^{3}}{3}

(as

s = 0

t = 0

)At

t = 2

s = -\dfrac{32}{3}

; at

t = 4

s = -\dfrac{16}{3}

First stage:

\dfrac{32}{3} + \left(-\dfrac{16}{3} + \dfrac{32}{3}\right) = 16

Second stage (velocity falls uniformly from

4

0

\dfrac{1}{2} \times \dfrac{8}{3} \times 4 = \dfrac{16}{3}

Total distance

= 16 + \dfrac{16}{3}

= \dfrac{64}{3}

Practice questions by skill

Sec 4 A-Math Kinematics worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Kinematics notes. Or practise it online with step-by-step help when you get one wrong.

Section A

(a)A particle moves in a straight line so that,

t

seconds after passing through a fixed point

O

, its velocity,

v

m/s, is given by

v = 9t^{2} - 8t

. Find its displacement from

O

when

t = 3

Answer (a)[3]

(b)A particle moves in a straight line so that,

t

seconds after passing through a fixed point

O

, its acceleration,

a

m/s², is given by

a = 4t + 3

. Its initial velocity is

-3

m/s. Find its velocity when

t = 2

Answer (b)[3]

Section B

A particle moves in a straight line so that,

t

seconds after passing through a fixed point

O

, its velocity,

v

m/s, is given by

v = 12e^{-\frac{t}{2}} - 2

(a)Find the initial velocity of the particle.

Answer (a)m/s[1]

(b)Find the initial acceleration of the particle.

Answer (b)m/s²[2]

(c)Find the exact value of

t

at which the particle is instantaneously at rest.

Answer (c)[2]

(d)Find the displacement of the particle from

O

when it is instantaneously at rest, correct to 3 significant figures.

Answer (d)m[2]

(e)Find the total distance travelled by the particle in the first

4

seconds, correct to 3 significant figures.

Answer (e)m[3]

Answers and worked solutions

Question 1

(a)Answer:

= 45

m[3]

s = \dfrac{9t^{3}}{3} - \dfrac{8t^{2}}{2} + c

= 3t^{3} - 4t^{2} + c

0 = 0 + c

c = 0

s = 3(3)^{3} - 4(3)^{2}

= 45

(b)Answer:

= 11

m/s[3]

v = \dfrac{4t^{2}}{2} + 3t + c

= 2t^{2} + 3t + c

-3 = 0 + c

c = -3

v = 2(2)^{2} + 3(2) - 3

= 11

m/s

Question 2

(a)Answer:

= 308

m[3]

3t^{2} - 42t + 120 = 0

3(t - 4)(t - 10) = 0

t = 4

t = 10

It turns back at

t = 4

s = \dfrac{3t^{3}}{3} - \dfrac{42t^{2}}{2} + 120t

= t^{3} - 21t^{2} + 120t

When

t = 4

s = 208

When

t = 9

s = 108

\text{Distance} = \left|208 - 0\right| + \left|108 - 208\right|

\text{Distance} = 208 + 100

= 308

(b)Answer:

= \dfrac{59}{3}

m[3]

t^{2} - 8t + 12 = 0

(t - 2)(t - 6) = 0

t = 2

t = 6

It turns back at

t = 2

s = \dfrac{t^{3}}{3} - \dfrac{8t^{2}}{2} + 12t

= \dfrac{1}{3}t^{3} - 4t^{2} + 12t

When

t = 2

s = \dfrac{32}{3}

When

t = 5

s = \dfrac{5}{3}

\text{Distance} = \left|\dfrac{32}{3} - 0\right| + \left|\dfrac{5}{3} - \dfrac{32}{3}\right|

\text{Distance} = \dfrac{32}{3} + 9

= \dfrac{59}{3}

Question 3

(a)Answer:

= -12

m/s²[3]

v = \dfrac{ds}{dt}

v = 3t^{2} - 12(2)t + 36(1)

= 3t^{2} - 24t + 36

a = \dfrac{dv}{dt}

a = 3(2)t - 24(1)

= 6t - 24

a = 6(2) - 24

= -12

m/s²

(b)Answer:

t = 1

t = 7

[3]

v = \dfrac{ds}{dt}

v = 2(3)t^{2} - 24(2)t + 42(1)

= 6t^{2} - 48t + 42

At rest:

v = 0

6t^{2} - 48t + 42 = 0

t^{2} - 8t + 7 = 0

(t - 1)(t - 7) = 0

t = 1

t = 7

Question 4

(a)Answer:

10

m/s[1]

When

t = 0

v = 12e^{0} - 2

= 10

m/s

(b)Answer:

-6

m/s²[2]

a = \dfrac{dv}{dt}

= -6e^{-\frac{t}{2}}

When

t = 0

a = -6

m/s²

(c)Answer:

t = 2\ln 6

[2]

12e^{-\frac{t}{2}} - 2 = 0

e^{-\frac{t}{2}} = \dfrac{1}{6}

-\frac{t}{2} = -\ln 6

t = 2\ln 6

(d)Answer:

12.8

m[2]

s = \int \left(12e^{-\frac{t}{2}} - 2\right) dt

= -24e^{-\frac{t}{2}} - 2t + c

s = 0

when

t = 0

, so

c = 24

s = 24 - 24e^{-\frac{t}{2}} - 2t

When

t = 2\ln 6

s = 24 - 4 - 2\left(2\ln 6\right)

= 12.8

m (3 s.f.)

(e)Answer:

12.9

m[3]

The particle is at rest at

t = 2\ln 6 \approx 3.584

, and moves back towards

O

after that.When

t = 4

s = 24 - 24e^{-2} - 8

s = 12.752

Distance

= 12.833 + (12.833 - 12.752)

= 12.9

m (3 s.f.)

Question 5

(a)Answer:

v = -12 + 8t - t^{2}

[2]

v = \int (8 - 2t) \, dt

= 8t - t^{2} + c_{1}

v = -12

when

t = 0

, so

c_{1} = -12

v = -12 + 8t - t^{2}

(b)Answer:

t = 2

s[2]

-12 + 8t - t^{2} = 0

t^{2} - 8t + 12 = 0

(t - 2)(t - 6) = 0

t = 2

t = 6

; the first is

t = 2

(and

t = 6

is after

T

)

(c)Answer:

T = 4

s[1]

The velocity is greatest when

a = 0

8 - 2T = 0

T = 4

(d)Answer:

4

m/s[1]

v = -12 + 8(4) - (4)^{2}

= 4

m/s

(e)Answer:

t = \dfrac{20}{3}

s[2]

After

t = 4

v = 4 - 1.5(t - 4)

v = 0

when

t - 4 = \dfrac{4}{1.5}

t - 4 = \dfrac{8}{3}

t = \dfrac{20}{3}

(f)Answer:

\dfrac{64}{3}

m[4]

s = \int (-12 + 8t - t^{2}) \, dt

\phantom{s} = -12t + 4t^{2} - \dfrac{t^{3}}{3}

(as

s = 0

t = 0

)At

t = 2

s = -\dfrac{32}{3}

; at

t = 4

s = -\dfrac{16}{3}

First stage:

\dfrac{32}{3} + \left(-\dfrac{16}{3} + \dfrac{32}{3}\right) = 16

Second stage (velocity falls uniformly from

4

0

\dfrac{1}{2} \times \dfrac{8}{3} \times 4 = \dfrac{16}{3}

Total distance

= 16 + \dfrac{16}{3}

= \dfrac{64}{3}