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Investigating Forces & Motion
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Investigating Forces and Motion (1998)(Granada Learning).iso
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topic3
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1998-02-16
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61 lines
[general]
[page1]
type:1
caption:\
These five velocity-time graphs record the motion of objects \
travelling for 4.0 seconds with uniform acceleration. Drag the arrows \
to match the values of the displacement, initial velocity, final \
velocity, or acceleration to the graphs.<p>
feedback:\
Correct. You can read the initial velocity, <i>u</i>, and final \
velocity, <i>v</i>, directly from the graphs. The acceleration, \
<i>a</i>, is the gradient of the graph. The increase in displacement, \
<i>s</i>, is the area under the graph.<p>
source:3ex1d,3ex1c,3ex1e,3ex1b,3ex1a
target:3ex1a1,3ex1b1,3ex1c1,3ex1d1,3ex1e1
[page2]
type:0
caption:\
<img src="3ex2" align=center><p>\
Many problems of uniformly accelerated motion involve gravity. For \
example, suppose that you throw a ball vertically upwards with an \
initial velocity of 10 m/s. How high does the ball go? How long is it \
before it returns to your hand? What is its velocity when it returns \
to your hand? Assume that the acceleration due to gravity is 10 \
m/s<SUP>2</SUP>.<p>\
With this kind of problem, it is always important to have a sign \
convention.<p>\
Sign convention: up is positive.<p>\
When the ball reaches its highest point, its velocity becomes zero for \
an instant.<p>\
For the first part of the ball's motion,<p>\
<I>u</I> = 10 m/s<p>\
<I>v</I> = 0.0 m/s<p>\
a = -10 m/s<SUP>2</SUP><p>\
(Because the gravity always acts in a downward direction, the value of \
the acceleration due to gravity will always be -10 m/s<SUP>2</SUP>, \
regardless of whether the ball is moving up or down.)<p>\
<I>s</I> = ?<p>\
<I>t</I> = ?<p>\
Use <I>v </I>= <I>u</I> + <I>at<p>\
</I><p>\
0.0 = 10 - 10<I>t<p>\
</I><p>\
Therefore,<p>\
10<I>t</I> = 10<p>\
<I>t</I> = 1.0 s<p>\
Use: <I>s</I> = <I>ut</I> + ½ <I>at</I><SUP>2 </sup><p>\
<I>s</I> = 10 x 1.0 + ½ x (-10) x 1.0<SUP>2</SUP><p>\
= 10 - 5.0<p>\
= 5.0 m<p>\
The second part of the ball's flight is exactly the reverse of the \
first part, and the ball takes the same time to fall as it did to \
rise.<p>\
Therefore, the total time of flight = 1.0 + 1.0 = 2.0 s<p>\
When the ball returns to your hand, the ball will be travelling at the \
same speed as it left it, but in the opposite direction. Its velocity \
will therefore be -10 m/s.<p>