http://sgforums.com/forums/2297/topics/316061 en-US 60 <p>for the 3rd question. deceleration... that's gravity so if we take the positive y direction as "upwards" acceleration would be -9.81 m/s^2&nbsp; or deceleration would be 9.81m/s^2 down.</p> Thu, 01 May 2008 17:11:48 +0800 sgforums.com:2297:316061:8053357 WoAiMeiMei http://sgforums.com/forums/2297/topics/316061 <p>For the first one, the answer is 10/3600 x 36 = .1km = 100m (you are correct)</p> <p>2. s = (1/2)(u+v)(t)<br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= (1/2)(10+25)(15)<br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =1/2 * 35 * 15<br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= 262.5</p> <p>Youll need to&nbsp;factor in the initial velocity too. If it starts at rest, it would be 187.5 but as it starts at 10ms-1, you would have to consider that when you work.</p> <p>3. dunch noes. Ask ur teacher!&nbsp;I think that it is correct though. As it states that the upwards velocity is 20m/s, the upwards direction is postive.</p> <p>Gah. someone beat me to the answers.</p> Wed, 30 Apr 2008 23:23:40 +0800 sgforums.com:2297:316061:8051466 jayh272416 http://sgforums.com/forums/2297/topics/316061 <p>1.</p> <p>d = s*t<br /> = 10m/s * 10s<br /> = 100m</p> <p>2.</p> <p>d = area under graph<br /> = area of rectangle + area of triangle<br /> = 10m/s * 15s + 0.5 * 15m/s * 15s<br /> = 262.5m</p> <p>for qn 2, if u draw out correctly, the graph should consist of a rectangle and a triangle</p> <p>3.</p> <p>it is correct to use deceleration (which is -ve acceleration), which is the rate of decrease of velocity(?)<br /> i dunno how to answer this qn &gt;.&gt;</p> Wed, 30 Apr 2008 23:20:05 +0800 sgforums.com:2297:316061:8051451 wishboy http://sgforums.com/forums/2297/topics/316061 <p>A train moves with a uniform velocity of 36km/h for 10 s. calculate the distance travelled.my answer's 100m. but my answer sheet states 6km. this is impossible because if you travel 6km in 10 secs, you'd cover 36km in 1 minute! you needn't take 1 hour. please check whether my answer is correct.</p> <p>next, a car has a velocity of 10m/s. it now accelerates at 1m/s2 for 1/4min(15secs). find the distance travelled in this time and final speed of the car. since they're asking for this distance travelled during this acceleration period, then the car would have a&nbsp;straight line&nbsp;in a speed/time graph. to find distance, then we'd use the area under the speed/.time grpah to find distance. final speed is 25m/s so the distance can be found be 0.5 X 15 X 25. this gives me 187.5 m but the answer writes 262.5m.</p> <p>&nbsp;</p> <p>the next question goes something like this</p> <p>a tennis ball is hit vertically upwards with a velocity of 20m/s.</p> <p>what is its <strong>deceleration</strong> as it moves upwards? is it correct to use the term 'deceleration'? i suppose so cuz objects accelerate at a constant speed&nbsp;when they fall without air resistance. but could someone clarify all these?</p> <p>&nbsp;</p> <p>thanks!</p> Wed, 30 Apr 2008 23:06:45 +0800 sgforums.com:2297:316061:8051406 bonkysleuth http://sgforums.com/forums/2297/topics/316061