<li>A car located 150 m from a street corner moves away from the corner at the uniform rate of 30. m/s. After maintaining this velocity for 10.s, it decelerates to a stop in 5.0 s. It then rests for another 5.0 s, then accelerates away from the corner at the rate of 2.0 m/s^2 for 10. s. Construct a celocity-time graph representing the motion of the car for the entire trip.</li>
</ol>
<p>[It may to difficult to construct a graph on CC, but please try your best to construct a table so I can use the data to make the graph.]</p>
<li>What is the magnitude of the vector sum of the two concurrent corces represented in the diagram? (1) 2.5 N (2) 3.5 N (3) 3.0 N (4) 4.0 N</li>
</ol>
<p>[Please explain answer to 51.]</p>
<pre><code> /l
l
l
l 2.0 N
l
l___________
1.5 N /
</code></pre>
<li>Two concurrent forces act at right angles to each other. If one of the forces is 40 newtons and the resultant of the two forces is 50 newtons the magnitude of the other force must be (1) 10 N (2) 20 N (3) 30 N (4) 40 N</li>
</ol>
<p>[Please explain answer to 54.]</p>
<li>If two 10.-newton concurrent forces have a resultant of zero, the angle between the forces must be (1) 0 degrees (2) 45 degrees (3) 90 degrees (4) 180 degrees</li>
</ol>
<p>45.)Hmmm, using y as distance and x as time. I think this is how it would work. This is gonna be tough to explain online.</p>
<p>For (x->1:10), you want to have a line that has a slope at 30.</p>
<p>The first points you want to have are (0,0) & (300, 10) and connect those points with a straight line. Then you would want to have the line become less steep for the next five seconds. When x=15, draw a straight horizontal line till x=20. Then for the last ten seconds, draw a line with an increasing slope.</p>
<p>That would be my guess.
NOTE:I'm really really bad at explaining things like this.</p>
<p>51.) In order to do a vector sum you have to calculate the square root of the sum of the squares of magnitudes. In this case it would be.</p>
<p>(2^2 + 1.5^2)^.5, or in other words 2.5.
Choice (1)</p>
<p>54, I think would be similar to 51. One thing you know is that there are two forces which make a right angle which you can use the Pythagorean Theorem for.
This would be (50^2-40^2)^.5 or in other words 30, choice (3).</p>
<p>This is possible beacue 50 is the resultant and 40 is one of the forces which form a right angle.</p>
<p>55.) (4.) When two equal forces are in opposite directions (or in a 180 degree angle), they cancel out to zero.</p>
<p>One force would be 10 Newtons and the other force would be -10 Newtons. 10-10=0.</p>
<ol>
<li>Construct a vector ing a velocity of 80. m/s directed at 60 degrees N of E. State the scale upon which the vector is based and construct the vector to scale. Then resolve the vector by construction and measurement into an easterly and northerly component. State the magnitude and direction of each of these compoent velocities.</li>
</ol>
<p>[Please describe how this diagram is drawn and what I have to do. I have no idea what the term "N of E' means. Does that mean Northeast 60 degrees according to the problem? So that it is pointing somewhere at 2 o'clock?] Please help with this.</p>
<ol>
<li>'60 degrees N of E' means the vector is pointing at 1 o'clock, not 2. Each 'hour' on the clock face represents (360/12) = a 30 degree shift. </li>
</ol>
<p>Think of a right-angled triangle with the hypotenuse at an angle of 60 degrees from the base, and the base = 1. You will then have height = sqrt(3) and hypotenuse = sqrt( 1 + 3) = 2. Now multiply each of these lengths by 40; the new lengths will be base=40, height = 40sqrt(3), hypotenuse = 80. Does this help?</p>
<p>The horizontal component (pointing East) has magnitude 40, the vertical (North) component has magnitude 40 sqrt(3). The resulting vector has a magnitude of 80.</p>
<p>If a vector component must be at right angles to all other components(I think this is part of the definition?), then any force in 3-D space can be resolved into at most 3 components. However, if components need not be at right angles to each other, then an infinite #components are possible.</p>