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Physics - Kinematics. Urgent help needed!

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  • bonkysleuth's Avatar
    188 posts since Mar '07
    • 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.

      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 straight line 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.

       

      the next question goes something like this

      a tennis ball is hit vertically upwards with a velocity of 20m/s.

      what is its deceleration as it moves upwards? is it correct to use the term 'deceleration'? i suppose so cuz objects accelerate at a constant speed when they fall without air resistance. but could someone clarify all these?

       

      thanks!

      Edited by bonkysleuth 30 Apr `08, 11:08PM
  • wishboy's Avatar
    1,131 posts since Aug '05
    • 1.

      d = s*t
      = 10m/s * 10s
      = 100m

      2.

      d = area under graph
      = area of rectangle + area of triangle
      = 10m/s * 15s + 0.5 * 15m/s * 15s
      = 262.5m

      for qn 2, if u draw out correctly, the graph should consist of a rectangle and a triangle

      3.

      it is correct to use deceleration (which is -ve acceleration), which is the rate of decrease of velocity(?)
      i dunno how to answer this qn >.>

  • jayh272416's Avatar
    305 posts since Aug '07
    • For the first one, the answer is 10/3600 x 36 = .1km = 100m (you are correct)

      2. s = (1/2)(u+v)(t)
             = (1/2)(10+25)(15)
              =1/2 * 35 * 15
              = 262.5

      Youll need to 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.

      3. dunch noes. Ask ur teacher! I think that it is correct though. As it states that the upwards velocity is 20m/s, the upwards direction is postive.

      Gah. someone beat me to the answers.

      Edited by jayh272416 30 Apr `08, 11:24PM
  • WoAiMeiMei's Avatar
    146 posts since Jul '07
    • 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  or deceleration would be 9.81m/s^2 down.

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