19. Exercises - Revision history

Ian at 17:18, 27 March 2011

Ian — Sun, 27 Mar 2011 17:18:32 GMT

←Older revision		Revision as of 17:18, 27 March 2011
Line 80:		Line 80:
	seconds is given by		seconds is given by

-	<math>a=\frac{4t}{5}</math>	+	<math>a=\frac{2t}{5}</math>
	for		for
	<math>0\le t\le 20.</math>		<math>0\le t\le 20.</math>

Ian at 13:37, 11 October 2010

Ian — Mon, 11 Oct 2010 13:37:54 GMT

←Older revision		Revision as of 13:37, 11 October 2010
Line 73:		Line 73:
	A car, of mass 1000 kg, accelerates for 20 seconds along a straight road. The graph below shows how the resultant force on the car varies with time as it accelerates.		A car, of mass 1000 kg, accelerates for 20 seconds along a straight road. The graph below shows how the resultant force on the car varies with time as it accelerates.

-	[[Image:E19.4.GIF]]	+	[[Image:E19.4new.GIF]]

	a) Show that the acceleration,		a) Show that the acceleration,

Ian at 09:23, 11 October 2010

Ian — Mon, 11 Oct 2010 09:23:38 GMT

←Older revision		Revision as of 09:23, 11 October 2010
Line 59:		Line 59:
	is given by:		is given by:

-	<math>a=8-\frac{8t}{10}</math>.	+	<math>a=8-\frac{4t}{5}</math>.

	b) Find the speed of the dragster when it stops accelerating at the end of the 10 seconds.		b) Find the speed of the dragster when it stops accelerating at the end of the 10 seconds.

Ian at 11:45, 8 March 2010

Ian — Mon, 08 Mar 2010 11:45:14 GMT

←Older revision		Revision as of 11:45, 8 March 2010
Line 2:		Line 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Not selected tab\|[[19. ~~Variable~~ acceleration II\|Theory]]}}	+	{{Not selected tab\|[[19. Motion with variable acceleration II\|Theory]]}}
	{{Selected tab\|[[19. Exercises\|Exercises]]}}		{{Selected tab\|[[19. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|

Ian at 19:08, 19 November 2009

Ian — Thu, 19 Nov 2009 19:08:55 GMT

←Older revision		Revision as of 19:08, 19 November 2009
Line 163:		Line 163:
	<div class="ovning">		<div class="ovning">

		+	An aeroplane moves so that at time <math>t</math>
		+	seconds its acceleration,
		+	<math>\mathbf{a}</math>
		+	<math>\text{m}{{\text{s}}^{-2}}</math>
		+	is given by:

		+	<math>\mathbf{a}=-2\mathbf{j}</math>
		+
		+	When
		+	<math>t=\text{ }0</math>
		+	the velocity of the aeroplane is (40<math>\mathbf{i}</math>) <math>\text{m}{{\text{s}}^{-1}}</math>and its position vector is
		+	(400<math>\mathbf{j}</math>) metres.
		+	The unit vectors <math>\mathbf{i}</math> and <math>\mathbf{j}</math> are directed east and north respectively.
		+
		+	a) Find the velocity of the aeroplane at time <math>t</math>.
		+
		+	b) Find the position vector of the particle at time <math>t</math>.
		+
		+	c) Find the time when the aeroplane is travelling south east.
		+
		+	d) Find the time when the aeroplane is due east of the origin.

	</div>{{#NAVCONTENT:Answer a\|Answer 19.8a\|Answer b\|Answer 19.8b\|Answer c\|Answer 19.8c\|Answer d\|Answer 19.8d\|Solution a\|Solution 19.8a\|Solution b\|Solution 19.8b\|Solution c\|Solution 19.8c\|Solution d\|Solution 19.8d}}		</div>{{#NAVCONTENT:Answer a\|Answer 19.8a\|Answer b\|Answer 19.8b\|Answer c\|Answer 19.8c\|Answer d\|Answer 19.8d\|Solution a\|Solution 19.8a\|Solution b\|Solution 19.8b\|Solution c\|Solution 19.8c\|Solution d\|Solution 19.8d}}

Ian at 18:53, 19 November 2009

Ian — Thu, 19 Nov 2009 18:53:53 GMT

←Older revision		Revision as of 18:53, 19 November 2009
Line 133:		Line 133:
	<div class="ovning">		<div class="ovning">

		+	A particle has mass 4 kg. It moves so that the resultant force,
		+	<math>\mathbf{F}</math>
		+	N, acting on it at time
		+	<math>t</math>
		+	seconds is given by:
		+
		+	<math>\mathbf{F}=12t\mathbf{i}+(6-5t)\mathbf{j}</math>
		+
		+	The unit vectors <math>\mathbf{i}</math> and <math>\mathbf{j}</math> are perpendicular.
		+
		+	a) Find the acceleration of the particle.
		+
		+	b) Find the velocity of the particle at time
		+	<math>t</math>
		+	if it has an initial velocity
		+	of -8<math>\mathbf{i}</math> <math>\text{m}{{\text{s}}^{-1}}</math>.
		+
		+	c) Given that the initial position of the particle is (<math>90\mathbf{i}+20\mathbf{j}</math>) metres, find an expression for the position vector of the particle at time
		+	<math>t</math>.
		+
		+	d) Find the distance of the particle from the origin when
		+	<math>t=\text{ 2}0</math>.

Ian at 18:43, 19 November 2009

Ian — Thu, 19 Nov 2009 18:43:41 GMT

←Older revision		Revision as of 18:43, 19 November 2009
Line 128:		Line 128:

	</div>{{#NAVCONTENT:Answer a\|Answer 19.6a\|Answer b\|Answer 19.6b\|Solution a\|Solution 19.6a\|Solution b\|Solution 19.6b}}		</div>{{#NAVCONTENT:Answer a\|Answer 19.6a\|Answer b\|Answer 19.6b\|Solution a\|Solution 19.6a\|Solution b\|Solution 19.6b}}
		+
		+
		+	===Exercise 19.7===
		+	<div class="ovning">
		+
		+
		+
		+	</div>{{#NAVCONTENT:Answer a\|Answer 19.7a\|Answer b\|Answer 19.7b\|Answer c\|Answer 19.7c\|Answer d\|Answer 19.7d\|Solution a\|Solution 19.7a\|Solution b\|Solution 19.7b\|Solution c\|Solution 19.7c\|Solution d\|Solution 19.7d}}
		+
		+
		+	===Exercise 19.8===
		+	<div class="ovning">
		+
		+
		+
		+	</div>{{#NAVCONTENT:Answer a\|Answer 19.8a\|Answer b\|Answer 19.8b\|Answer c\|Answer 19.8c\|Answer d\|Answer 19.8d\|Solution a\|Solution 19.8a\|Solution b\|Solution 19.8b\|Solution c\|Solution 19.8c\|Solution d\|Solution 19.8d}}

Ian at 18:25, 19 November 2009

Ian — Thu, 19 Nov 2009 18:25:31 GMT

←Older revision		Revision as of 18:25, 19 November 2009
Line 110:		Line 110:

	b) Find the distance that the car travels during the 40 seconds.		b) Find the distance that the car travels during the 40 seconds.
-

	</div>{{#NAVCONTENT:Answer a\|Answer 19.5a\|Answer b\|Answer 19.5b\|Solution a\|Solution 19.5a\|Solution b\|Solution 19.5b}}		</div>{{#NAVCONTENT:Answer a\|Answer 19.5a\|Answer b\|Answer 19.5b\|Solution a\|Solution 19.5a\|Solution b\|Solution 19.5b}}
		+
		+
		+	===Exercise 19.6===
		+	<div class="ovning">
		+
		+	A car is moving along a straight road. The velocity of the car at time <math>t</math> seconds is
		+	<math>\left( 3{{t}^{2}}+2t+5 \right)</math>
		+	<math>\text{m}{{\text{s}}^{-2}}</math>.
		+
		+	a) Find an expression for the acceleration of the car at time <math>t</math>.
		+
		+	b) Find the total distance travelled by the car when
		+	<math>t=\text{ 1}0</math>
		+
		+
		+	</div>{{#NAVCONTENT:Answer a\|Answer 19.6a\|Answer b\|Answer 19.6b\|Solution a\|Solution 19.6a\|Solution b\|Solution 19.6b}}

Ian at 18:18, 19 November 2009

Ian — Thu, 19 Nov 2009 18:18:17 GMT

←Older revision		Revision as of 18:18, 19 November 2009
Line 90:		Line 90:

	c) Find the distance travelled by the car in the 20 second period.		c) Find the distance travelled by the car in the 20 second period.
-

	</div>{{#NAVCONTENT:Answer b\|Answer 19.4b\|Answer c\|Answer 19.4c\|Solution a\|Solution 19.4a\|Solution b\|Solution 19.4b\|Solution c\|Solution 19.4c}}		</div>{{#NAVCONTENT:Answer b\|Answer 19.4b\|Answer c\|Answer 19.4c\|Solution a\|Solution 19.4a\|Solution b\|Solution 19.4b\|Solution c\|Solution 19.4c}}
		+
		+
		+	===Exercise 19.5===
		+	<div class="ovning">
		+
		+	A van slows down from a speed of 20 <math>\text{m}{{\text{s}}^{-1}}</math>
		+	to 5 <math>\text{m}{{\text{s}}^{-1}}</math>
		+	in 40 seconds. At time <math>t</math>
		+	seconds the acceleration, <math>a</math>
		+	<math>\text{m}{{\text{s}}^{-2}}</math>
		+	is given by
		+	<math>a=-kt</math>, where
		+	<math>k</math> is a constant.
		+
		+	a) Find
		+	<math>k</math>.
		+
		+	b) Find the distance that the car travels during the 40 seconds.
		+
		+
		+	</div>{{#NAVCONTENT:Answer a\|Answer 19.5a\|Answer b\|Answer 19.5b\|Solution a\|Solution 19.5a\|Solution b\|Solution 19.5b}}

Ian at 18:10, 19 November 2009

Ian — Thu, 19 Nov 2009 18:10:36 GMT

←Older revision		Revision as of 18:10, 19 November 2009
Line 65:		Line 65:
	c) Find the distance that the dragster travels in the 10 seconds.		c) Find the distance that the dragster travels in the 10 seconds.

		+	</div>{{#NAVCONTENT:Answer b\|Answer 19.3b\|Answer c\|Answer 19.3c\|Solution a\|Solution 19.3a\|Solution b\|Solution 19.3b\|Solution c\|Solution 19.3c}}


-	</div>{{#NAVCONTENT:Answer b\|Answer 19.3b\|Answer c\|Answer 19.3c\|Solution a\|Solution 19.3a\|Solution b\|Solution 19.3b\|Solution c\|Solution 19.3c}}	+	===Exercise 19.4===
		+	<div class="ovning">
		+
		+	A car, of mass 1000 kg, accelerates for 20 seconds along a straight road. The graph below shows how the resultant force on the car varies with time as it accelerates.
		+
		+	[[Image:E19.4.GIF]]
		+
		+	a) Show that the acceleration,
		+	<math>a</math>
		+	<math>\text{m}{{\text{s}}^{-2}}</math>, at time <math>t</math>
		+	seconds is given by
		+
		+	<math>a=\frac{4t}{5}</math>
		+	for
		+	<math>0\le t\le 20.</math>
		+
		+
		+	b) Given that the velocity of the car is 8 <math>\text{m}{{\text{s}}^{-1}}</math>
		+	when it starts to accelerate, find an expression for the velocity of the car at time <math>t</math>
		+	seconds.
		+
		+	c) Find the distance travelled by the car in the 20 second period.
		+
		+
		+	</div>{{#NAVCONTENT:Answer b\|Answer 19.4b\|Answer c\|Answer 19.4c\|Solution a\|Solution 19.4a\|Solution b\|Solution 19.4b\|Solution c\|Solution 19.4c}}

←Older revision		Revision as of 11:45, 8 March 2010
Line 2:		Line 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Not selected tab\|[[19. ~~Variable~~ acceleration II\|Theory]]}}	+	{{Not selected tab\|[[19. Motion with variable acceleration II\|Theory]]}}
	{{Selected tab\|[[19. Exercises\|Exercises]]}}		{{Selected tab\|[[19. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|