6. Kinematics in one dimension - Revision history

Ian at 14:50, 11 April 2011

2011-04-11T14:50:53Z

←Older revision		Revision as of 14:50, 11 April 2011
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	{{Selected tab\|[[6. Kinematics in one dimension\|Theory]]}}		{{Selected tab\|[[6. Kinematics in one dimension\|Theory]]}}
	{{Not selected tab\|[[6. Exercises\|Exercises]]}}		{{Not selected tab\|[[6. Exercises\|Exercises]]}}
		+	{{Not selected tab\|[[6. Video\|Video]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}
-
-

	== '''Key Points''' ==		== '''Key Points''' ==

Ian at 12:03, 27 March 2011

2011-03-27T12:03:17Z

←Older revision		Revision as of 12:03, 27 March 2011
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	<math>\begin{align}		<math>\begin{align}
-	& 0=10+(-0\textrm{.}8)t \\	+	& 0=18+(-1\textrm{.}2)t \\
-	& 0=10-0\textrm{.}8t \\	+	& 0=18-1\textrm{.}2t \\
-	& t=\frac{10}{0\textrm{.}8}=~~12\textrm{.}5~~\text{ s} \\	+	& t=\frac{18}{1\textrm{.}2}=15\text{ s} \\
	\end{align}</math>		\end{align}</math>

Ian at 16:34, 4 March 2011

2011-03-04T16:34:59Z

←Older revision		Revision as of 16:34, 4 March 2011
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	<math>\begin{align}		<math>\begin{align}
	& {{0}^{2}}={{18}^{2}}+2\times (-1\textrm{.}2)s \\		& {{0}^{2}}={{18}^{2}}+2\times (-1\textrm{.}2)s \\
-	& 324-2.4s=0 \\	+	& 324-2\textrm{.}4s=0 \\
	& s=\frac{324}{2\textrm{.}4}=135\text{ m} \\		& s=\frac{324}{2\textrm{.}4}=135\text{ m} \\
	\end{align}</math>		\end{align}</math>

Ian at 18:54, 10 March 2010

2010-03-10T18:54:00Z

←Older revision		Revision as of 18:54, 10 March 2010
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	b) Calculate the acceleration of the car on each on each stage of its journey.		b) Calculate the acceleration of the car on each on each stage of its journey.
		+

	[[Image:ex6.1fig1test1.gif]]		[[Image:ex6.1fig1test1.gif]]

Ian at 18:51, 10 March 2010

2010-03-10T18:51:57Z

←Older revision		Revision as of 18:51, 10 March 2010
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	b) Calculate the acceleration of the car on each on each stage of its journey.		b) Calculate the acceleration of the car on each on each stage of its journey.

-	[[Image:ex6.1fig1test1.~~jpg~~]]	+	[[Image:ex6.1fig1test1.gif]]

	'''Solution'''		'''Solution'''

Ian at 11:31, 8 March 2010

2010-03-08T11:31:35Z

←Older revision		Revision as of 11:31, 8 March 2010
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	\|-		\|-
	\| <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math>		\| <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math>
-	\|valign="top"\|<math>s</math> = displacement	+	\|valign="top"\|<math>s</math> = displacement (from initial position)

	\|-		\|-

Ian at 15:32, 15 February 2010

2010-02-15T15:32:52Z

←Older revision		Revision as of 15:32, 15 February 2010
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	\| valign="top"\| <math>v</math> = final velocity		\| valign="top"\| <math>v</math> = final velocity
	\|-		\|-
-	\| <math>{{v}^{2}}={{u}^{2}}+2as</math>	+	\| <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math>
	\|valign="top"\| <math>a</math> = acceleration		\|valign="top"\| <math>a</math> = acceleration

	\|-		\|-
-	\| <math>s=ut+\frac{1}{2}a{{t}^{2}}</math>	+	\| <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math>
	\|valign="top"\|<math>s</math> = displacement		\|valign="top"\|<math>s</math> = displacement

Line 137:		Line 137:

	Using the equation		Using the equation
-	<math>{{v}^{2}}={{u}^{2}}+2as</math>	+	<math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math>
	gives:		gives:

Line 165:		Line 165:

	a) Using		a) Using
-	<math>s=ut+\frac{1}{2}a{{t}^{2}}</math>	+	<math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math>
	gives:		gives:

	<math>\begin{align}		<math>\begin{align}
-	& 10=0\times t+\frac{1}{2}\times 9\textrm{.}8{{t}^{2}} \\	+	& 10=0\times t+\frac{1}{2}\times 9\textrm{.}8{{t}^{\ 2}} \\
-	& 10=4\textrm{.}9{{t}^{2}} \\	+	& 10=4\textrm{.}9{{t}^{\ 2}} \\
	& t=\sqrt{\frac{10}{4\textrm{.}9}}=1\textrm{.}43\text{ s (to 3sf)} \\		& t=\sqrt{\frac{10}{4\textrm{.}9}}=1\textrm{.}43\text{ s (to 3sf)} \\
	\end{align}</math>		\end{align}</math>

	b) Using		b) Using
-	<math>{{v}^{2}}={{u}^{2}}+2as</math>	+	<math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math>
	gives:		gives:

	<math>\begin{align}		<math>\begin{align}
-	& {{v}^{2}}={{0}^{2}}+2\times 9\textrm{.}8\times 10 \\	+	& {{v}^{\ 2}}={{0}^{2}}+2\times 9\textrm{.}8\times 10 \\
	& v=\sqrt{196}=14\text{ m}{{\text{s}}^{\text{-1}}} \\		& v=\sqrt{196}=14\text{ m}{{\text{s}}^{\text{-1}}} \\
	\end{align}</math>		\end{align}</math>
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	Using		Using
-	<math>{{v}^{2}}={{u}^{2}}+2as</math>	+	<math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math>
	gives;		gives;

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	Using		Using
-	<math>s=ut+\frac{1}{2}a{{t}^{2}}</math>	+	<math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math>
	gives:		gives:

	<math>\begin{align}		<math>\begin{align}
-	& -2=9t+\frac{1}{2}\times (-10){{t}^{2}} \\	+	& -2=9t+\frac{1}{2}\times (-10){{t}^{\ 2}} \\
-	& -2=9t-5{{t}^{2}} \\	+	& -2=9t-5{{t}^{\ 2}} \\
-	& 5{{t}^{2}}-9t-2=0 \\	+	& 5{{t}^{\ 2}}-9t-2=0 \\
	& (5t+1)(t-2)=0 \\		& (5t+1)(t-2)=0 \\
	& t=-\frac{1}{5}\text{ or }t=2 \\		& t=-\frac{1}{5}\text{ or }t=2 \\
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	a) Using		a) Using
-	<math>{{v}^{2}}={{u}^{2}}+2as</math>	+	<math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math>
	we can find the distance travelled:		we can find the distance travelled:

Ian at 11:41, 9 February 2010

2010-02-09T11:41:13Z

←Older revision		Revision as of 11:41, 9 February 2010
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	Stage 1:		Stage 1:
-	<math>a=\frac{6}{8}=0.75\text{ m}{{\text{s}}^{\text{-2}}}</math>	+	<math>a=\frac{6}{8}=0\textrm{.}75\text{ m}{{\text{s}}^{\text{-2}}}</math>

	Stage 2:		Stage 2:
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	& 35=10+50a \\		& 35=10+50a \\
	& 25=50a \\		& 25=50a \\
-	& a=\frac{25}{50}=0.5\text{ m}{{\text{s}}^{\text{-2}}} \\	+	& a=\frac{25}{50}=0\textrm{.}5\text{ m}{{\text{s}}^{\text{-2}}} \\
	\end{align}</math>		\end{align}</math>

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	<math>v=20</math>		<math>v=20</math>
	and		and
-	<math>a=0.5</math>	+	<math>a=0\textrm{.}5</math>
	.		.

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	<math>\begin{align}		<math>\begin{align}
-	& {{20}^{2}}={{10}^{2}}+2\times 0.5s \\	+	& {{20}^{2}}={{10}^{2}}+2\times 0\textrm{.}5s \\
	& 400=100+s \\		& 400=100+s \\
	& s=300\text{ m} \\		& s=300\text{ m} \\
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	<math>u=0</math>		<math>u=0</math>
	,		,
-	<math>a=9.8</math>	+	<math>a=9\textrm{.}8</math>
	and		and
	<math>s=10</math>		<math>s=10</math>
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	<math>\begin{align}		<math>\begin{align}
-	& 10=0\times t+\frac{1}{2}\times 9.8{{t}^{2}} \\	+	& 10=0\times t+\frac{1}{2}\times 9\textrm{.}8{{t}^{2}} \\
-	& 10=4.9{{t}^{2}} \\	+	& 10=4\textrm{.}9{{t}^{2}} \\
-	& t=\sqrt{\frac{10}{4.9}}=1.43\text{ s (to 3sf)} \\	+	& t=\sqrt{\frac{10}{4\textrm{.}9}}=1\textrm{.}43\text{ s (to 3sf)} \\
	\end{align}</math>		\end{align}</math>

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	<math>\begin{align}		<math>\begin{align}
-	& {{v}^{2}}={{0}^{2}}+2\times 9.8\times 10 \\	+	& {{v}^{2}}={{0}^{2}}+2\times 9\textrm{.}8\times 10 \\
	& v=\sqrt{196}=14\text{ m}{{\text{s}}^{\text{-1}}} \\		& v=\sqrt{196}=14\text{ m}{{\text{s}}^{\text{-1}}} \\
	\end{align}</math>		\end{align}</math>
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	& {{0}^{2}}={{9}^{2}}+2\times (-10)s \\		& {{0}^{2}}={{9}^{2}}+2\times (-10)s \\
	& 0=81-20s \\		& 0=81-20s \\
-	& s=\frac{81}{20}=4.05\text{ m} \\	+	& s=\frac{81}{20}=4\textrm{.}05\text{ m} \\
	\end{align}</math>		\end{align}</math>

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	<math>u=18</math>		<math>u=18</math>
	,		,
-	<math>a=-1.2</math>	+	<math>a=-1\textrm{.}2</math>
	and since the lorry comes to rest		and since the lorry comes to rest
	<math>v=0</math>		<math>v=0</math>
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	<math>\begin{align}		<math>\begin{align}
-	& {{0}^{2}}={{18}^{2}}+2\times (-1.2)s \\	+	& {{0}^{2}}={{18}^{2}}+2\times (-1\textrm{.}2)s \\
	& 324-2.4s=0 \\		& 324-2.4s=0 \\
-	& s=\frac{324}{2.4}=135\text{ m} \\	+	& s=\frac{324}{2\textrm{.}4}=135\text{ m} \\
	\end{align}</math>		\end{align}</math>

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	<math>\begin{align}		<math>\begin{align}
-	& 0=10+(-0.8)t \\	+	& 0=10+(-0\textrm{.}8)t \\
-	& 0=10-0.8t \\	+	& 0=10-0\textrm{.}8t \\
-	& t=\frac{10}{0.8}=12.5\text{ s} \\	+	& t=\frac{10}{0\textrm{.}8}=12\textrm{.}5\text{ s} \\
	\end{align}</math>		\end{align}</math>

Ian at 11:32, 9 February 2010

2010-02-09T11:32:58Z

←Older revision		Revision as of 11:32, 9 February 2010
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	'''[[Example 6.2]]'''		'''[[Example 6.2]]'''

-	As a car travels along a straight road its speed increases from 10 ms-1 to	+	As a car travels along a straight road its speed increases from 10 <math>\displaystyle \text{m}{{\text{s}}^{-1}}</math> to
-	35 ms-1 in 50 seconds.	+	35 <math>\displaystyle \text{m}{{\text{s}}^{-1}}</math> in 50 seconds.

	a) Calculate the distance travelled by the car.		a) Calculate the distance travelled by the car.
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	b) Find the acceleration of the car.		b) Find the acceleration of the car.

-	c) Find the distance travelled when the speed of the car is 20 ms-1.	+	c) Find the distance travelled when the speed of the car is 20 <math>\displaystyle \text{m}{{\text{s}}^{-1}}</math>.

	'''Solution'''		'''Solution'''
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	'''[[Example 6.4]]'''		'''[[Example 6.4]]'''

-	A ball is thrown vertically upwards at a speed of 9 ms-1, from a height of 2 m. Find the maximum height of the ball and the time it is in the air.	+	A ball is thrown vertically upwards at a speed of 9 <math>\displaystyle \text{m}{{\text{s}}^{-1}}</math>, from a height of 2 m. Find the maximum height of the ball and the time it is in the air.
-	Assume g = 10 ms-2.	+	Assume g = 10 <math>\displaystyle \text{m}{{\text{s}}^{-2}}</math>.

	'''Solution'''		'''Solution'''
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	'''[[Example 6.5]]'''		'''[[Example 6.5]]'''

-	A lorry is initially travelling at 18 ms-1, when the driver applies the brakes. Assume that the acceleration of the lorry remains constant at -1.2 ms-2 until it stops.	+	A lorry is initially travelling at 18 <math>\displaystyle \text{m}{{\text{s}}^{-1}}</math>, when the driver applies the brakes. Assume that the acceleration of the lorry remains constant at -1.2 <math>\displaystyle \text{m}{{\text{s}}^{-2}}</math> until it stops.
	(a) Find the distance that the lorry travels before stopping.		(a) Find the distance that the lorry travels before stopping.
	(b) Find the time that it takes for the lorry to stop.		(b) Find the time that it takes for the lorry to stop.

Ian at 11:19, 9 February 2010

2010-02-09T11:19:30Z

←Older revision		Revision as of 11:19, 9 February 2010
Line 55:		Line 55:

	The graph shows how the velocity of a car changes as it moves a short distance along a straight road.		The graph shows how the velocity of a car changes as it moves a short distance along a straight road.
-	(a) Find the total distance travelled by the car.
-	(b) Calculate the acceleration of the car on each on each stage of its journey.
-
-
-

		+	a) Find the total distance travelled by the car.

		+	b) Calculate the acceleration of the car on each on each stage of its journey.

	[[Image:ex6.1fig1test1.jpg]]		[[Image:ex6.1fig1test1.jpg]]
-

	'''Solution'''		'''Solution'''
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	Using the formula for the area of a trapezium gives:		Using the formula for the area of a trapezium gives:

-
	<math>s=\frac{1}{2}(12+26)\times 6=114\text{ m}</math>		<math>s=\frac{1}{2}(12+26)\times 6=114\text{ m}</math>
-

	Alternatively, breaking the area under the graph into three parts gives.		Alternatively, breaking the area under the graph into three parts gives.

-
	<math>\begin{align}		<math>\begin{align}
	& s=\left( \frac{1}{2}\times 8\times 6 \right)+\left( 12\times 6 \right)+\left( \frac{1}{2}\times 6\times 6 \right) \\		& s=\left( \frac{1}{2}\times 8\times 6 \right)+\left( 12\times 6 \right)+\left( \frac{1}{2}\times 6\times 6 \right) \\
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	& =114\text{ m}		& =114\text{ m}
	\end{align}</math>		\end{align}</math>
-

	(b) The acceleration of the car on each stage of the journey is given by the gradient of the graph.		(b) The acceleration of the car on each stage of the journey is given by the gradient of the graph.
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	Stage 1:		Stage 1:
	<math>a=\frac{6}{8}=0.75\text{ m}{{\text{s}}^{\text{-2}}}</math>		<math>a=\frac{6}{8}=0.75\text{ m}{{\text{s}}^{\text{-2}}}</math>
-

	Stage 2:		Stage 2:
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	Stage 3:		Stage 3:
	<math>a=-\frac{6}{6}=-1\text{ m}{{\text{s}}^{\text{-2}}}</math>		<math>a=-\frac{6}{6}=-1\text{ m}{{\text{s}}^{\text{-2}}}</math>
-
-


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	35 ms-1 in 50 seconds.		35 ms-1 in 50 seconds.

-	(a) Calculate the distance travelled by the car.	+	a) Calculate the distance travelled by the car.
-	(b) Find the acceleration of the car.	+
-	(c) Find the distance travelled when the speed of the car is 20 ms-1.	+	b) Find the acceleration of the car.
		+
		+	c) Find the distance travelled when the speed of the car is 20 ms-1.

	'''Solution'''		'''Solution'''
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	.		.

-	(a) Using	+	a) Using
	<math>s=\frac{1}{2}(u+v)t</math>		<math>s=\frac{1}{2}(u+v)t</math>
	gives:		gives:
-

	<math>s=\frac{1}{2}(10+35)\times 50=1125\text{ m}</math>		<math>s=\frac{1}{2}(10+35)\times 50=1125\text{ m}</math>

-		+	b) Using
-	(b) Using	+
	<math>v=u+at</math>		<math>v=u+at</math>
	gives:		gives:
-		+
-		+
	<math>\begin{align}		<math>\begin{align}
	& 35=10+50a \\		& 35=10+50a \\
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	\end{align}</math>		\end{align}</math>

-		+	c) We now have to work with
-	(c) We now have to work with	+
	<math>u=10</math>		<math>u=10</math>
	,		,
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	<math>{{v}^{2}}={{u}^{2}}+2as</math>		<math>{{v}^{2}}={{u}^{2}}+2as</math>
	gives:		gives:
-		+
-		+
	<math>\begin{align}		<math>\begin{align}
	& {{20}^{2}}={{10}^{2}}+2\times 0.5s \\		& {{20}^{2}}={{10}^{2}}+2\times 0.5s \\
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	A ball is dropped from a height of 10 m and falls until it hits the ground. Assume that no resistance forces act on the ball as it falls.		A ball is dropped from a height of 10 m and falls until it hits the ground. Assume that no resistance forces act on the ball as it falls.

-	(a) Find the time that it takes the ball to reach the ground.	+	a) Find the time that it takes the ball to reach the ground.
-	(b) Find the speed of the ball when it hits the ground.	+
		+	b) Find the speed of the ball when it hits the ground.

	'''Solution'''		'''Solution'''
-

	In this case we have		In this case we have
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	<math>s=10</math>		<math>s=10</math>

-		+	a) Using
-	(a) Using	+
	<math>s=ut+\frac{1}{2}a{{t}^{2}}</math>		<math>s=ut+\frac{1}{2}a{{t}^{2}}</math>
	gives:		gives:
-		+
-		+
	<math>\begin{align}		<math>\begin{align}
	& 10=0\times t+\frac{1}{2}\times 9.8{{t}^{2}} \\		& 10=0\times t+\frac{1}{2}\times 9.8{{t}^{2}} \\
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	\end{align}</math>		\end{align}</math>

-		+	b) Using
-	(b) Using	+
	<math>{{v}^{2}}={{u}^{2}}+2as</math>		<math>{{v}^{2}}={{u}^{2}}+2as</math>
	gives:		gives:
-		+
-		+
	<math>\begin{align}		<math>\begin{align}
	& {{v}^{2}}={{0}^{2}}+2\times 9.8\times 10 \\		& {{v}^{2}}={{0}^{2}}+2\times 9.8\times 10 \\
	& v=\sqrt{196}=14\text{ m}{{\text{s}}^{\text{-1}}} \\		& v=\sqrt{196}=14\text{ m}{{\text{s}}^{\text{-1}}} \\
	\end{align}</math>		\end{align}</math>
-


Line 209:		Line 190:

	'''Solution'''		'''Solution'''
-

	In this case		In this case
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	<math>{{v}^{2}}={{u}^{2}}+2as</math>		<math>{{v}^{2}}={{u}^{2}}+2as</math>
	gives;		gives;
-		+
-		+
	<math>\begin{align}		<math>\begin{align}
	& {{0}^{2}}={{9}^{2}}+2\times (-10)s \\		& {{0}^{2}}={{9}^{2}}+2\times (-10)s \\
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	& s=\frac{81}{20}=4.05\text{ m} \\		& s=\frac{81}{20}=4.05\text{ m} \\
	\end{align}</math>		\end{align}</math>
-

	But as the ball is thrown from a height of 2m, it reaches a maximum of 6.05 metres above ground level.		But as the ball is thrown from a height of 2m, it reaches a maximum of 6.05 metres above ground level.
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	<math>s=ut+\frac{1}{2}a{{t}^{2}}</math>		<math>s=ut+\frac{1}{2}a{{t}^{2}}</math>
	gives:		gives:
-

	<math>\begin{align}		<math>\begin{align}
Line 249:		Line 226:
	& t=-\frac{1}{5}\text{ or }t=2 \\		& t=-\frac{1}{5}\text{ or }t=2 \\
	\end{align}</math>		\end{align}</math>
-

	We need a positive time and the ball must be in the air for 2 seconds.		We need a positive time and the ball must be in the air for 2 seconds.
-


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	.		.

-	(a) Using	+	a) Using
	<math>{{v}^{2}}={{u}^{2}}+2as</math>		<math>{{v}^{2}}={{u}^{2}}+2as</math>
	we can find the distance travelled:		we can find the distance travelled:

-
	<math>\begin{align}		<math>\begin{align}
	& {{0}^{2}}={{18}^{2}}+2\times (-1.2)s \\		& {{0}^{2}}={{18}^{2}}+2\times (-1.2)s \\
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	& s=\frac{324}{2.4}=135\text{ m} \\		& s=\frac{324}{2.4}=135\text{ m} \\
	\end{align}</math>		\end{align}</math>
-

	Using		Using
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	we can find the time taken:		we can find the time taken:

-
	<math>\begin{align}		<math>\begin{align}
	& 0=10+(-0.8)t \\		& 0=10+(-0.8)t \\