標題:
Motion in one direction
發問:
題目圖片是網址中的67題http://books.google.com.hk/books?id=FRkFAAAAQBAJ&pg=PA54&lpg=PA54&dq=blue+car+of+length+4.52+m+is+moving+north+on+a+roadway+that+intersects+another+perpendicular+roadway+(Fig.+P2.67).+The+width+of+the+intersection+from+near&source=bl&ots=stTeEUrsTG&sig=pNzefIhCz34La_VGsGhFHCLqmx... 顯示更多 題目圖片是網址中的67題 http://books.google.com.hk/books?id=FRkFAAAAQBAJ&pg=PA54&lpg=PA54&dq=blue+car+of+length+4.52+m+is+moving+north+on+a+roadway+that+intersects+another+perpendicular+roadway+(Fig.+P2.67).+The+width+of+the+intersection+from+near&source=bl&ots=stTeEUrsTG&sig=pNzefIhCz34La_VGsGhFHCLqmx4&hl=zh-TW&sa=X&ei=7e1QUvfyF4LOiAeoqoG4Bg&ved=0CC8Q6AEwAA question: A blue car of length 4.52 m is moving north on a roadway that intersects another perpendicular roadway (Fig. P2.67). The width of the intersection from near edge to far edge is 28.0 m. The blue car has a constant acceleration of magnitude 2.10 m/s2 directed south. The time interval required for the nose of the blue car to move from the near (south) edge of the intersection to the north edge of the intersection is 3.10 s. (a) How far is the nose of the blue car from the south edge of the intersection when it stops? (b) For what time interval is any part of the blue car within the boundaries of the intersection? (c) A red car is at rest on the perpendicular intersecting roadway. As the nose of the blue car enters the intersection, the red car starts from rest and accelerates east at 5.60 m/s2. What is the minimum distance from the near (west) edge of the intersection at which the nose of the red car can begin its motion if it is to enter the intersection after the blue car has entirely left the intersection? (d) If the red car begins its motion at the position given by the answer to part (c), with what speed does it enter the intersection? 想了很久也不懂怎麼做,書本也沒有答案 麻煩師兄解答一下 thanks!
最佳解答:
(a) Consider the time when the blue car is passing the intersection, use equation: s = ut + (1/2)at^2 with s = 28 m, a = -2.1 m/s^2, t = 3.1 s, u =? hence, 28 = 3.1u + (1/2).(-2.1).(3.1^2) u = 12.29 m/s i.e. the speed of the blue car when its nose reaches the near (south) edge of the intersection is 12.29 m/s Then use equation: v^2 = u^2 + 2a.s with v = 0 m/s, u = 12.29 m/s, a = -2.1 m/s, s =? hence, 0 = 12.29^2 + 2.(-2.1)s s = 35.96 m Therefore, the nose of the blue car is 35.96 m from the south edge of the intersection when it stops. (b) The blue car needs to travel (28 + 4.52) m = 32.52 m from the south edge of the intersection in oder to completely clear from that intersection. Use equation: s = ut + (1/2)at^2 with s = 32.52 m, u = 12.29 m/s, a = -2.1 m/s^2, t = ? hence, 32.52 = 12.29t + (1/2).(-2.1)t^2 solve for t gives t = 7.663 s or -4.04 s (rejected) Therefore, the blue car needs 7.663 s to clear off the intersection. (c) Consider the red car. Use equation: s = ut + (1/2)at^2 with u = 0 m/s, t = 7.663 s, a = 5.6 m/s^2, s = ? hence, s = (1/2).(5.6).(7.663^2) m = 164.4 m The red car is at 164.4 m from the west edge of the intersection. (d) Use equation: v = u + at with u = 0 m/s, a = 5.6 m/s^2, t = 7.663 s, v =? hence, v = 5.6 x 7.663 m/s = 42.9 m/s The speed of the red car is 42.9 m/s
其他解答:
Motion in one direction
發問:
題目圖片是網址中的67題http://books.google.com.hk/books?id=FRkFAAAAQBAJ&pg=PA54&lpg=PA54&dq=blue+car+of+length+4.52+m+is+moving+north+on+a+roadway+that+intersects+another+perpendicular+roadway+(Fig.+P2.67).+The+width+of+the+intersection+from+near&source=bl&ots=stTeEUrsTG&sig=pNzefIhCz34La_VGsGhFHCLqmx... 顯示更多 題目圖片是網址中的67題 http://books.google.com.hk/books?id=FRkFAAAAQBAJ&pg=PA54&lpg=PA54&dq=blue+car+of+length+4.52+m+is+moving+north+on+a+roadway+that+intersects+another+perpendicular+roadway+(Fig.+P2.67).+The+width+of+the+intersection+from+near&source=bl&ots=stTeEUrsTG&sig=pNzefIhCz34La_VGsGhFHCLqmx4&hl=zh-TW&sa=X&ei=7e1QUvfyF4LOiAeoqoG4Bg&ved=0CC8Q6AEwAA question: A blue car of length 4.52 m is moving north on a roadway that intersects another perpendicular roadway (Fig. P2.67). The width of the intersection from near edge to far edge is 28.0 m. The blue car has a constant acceleration of magnitude 2.10 m/s2 directed south. The time interval required for the nose of the blue car to move from the near (south) edge of the intersection to the north edge of the intersection is 3.10 s. (a) How far is the nose of the blue car from the south edge of the intersection when it stops? (b) For what time interval is any part of the blue car within the boundaries of the intersection? (c) A red car is at rest on the perpendicular intersecting roadway. As the nose of the blue car enters the intersection, the red car starts from rest and accelerates east at 5.60 m/s2. What is the minimum distance from the near (west) edge of the intersection at which the nose of the red car can begin its motion if it is to enter the intersection after the blue car has entirely left the intersection? (d) If the red car begins its motion at the position given by the answer to part (c), with what speed does it enter the intersection? 想了很久也不懂怎麼做,書本也沒有答案 麻煩師兄解答一下 thanks!
最佳解答:
(a) Consider the time when the blue car is passing the intersection, use equation: s = ut + (1/2)at^2 with s = 28 m, a = -2.1 m/s^2, t = 3.1 s, u =? hence, 28 = 3.1u + (1/2).(-2.1).(3.1^2) u = 12.29 m/s i.e. the speed of the blue car when its nose reaches the near (south) edge of the intersection is 12.29 m/s Then use equation: v^2 = u^2 + 2a.s with v = 0 m/s, u = 12.29 m/s, a = -2.1 m/s, s =? hence, 0 = 12.29^2 + 2.(-2.1)s s = 35.96 m Therefore, the nose of the blue car is 35.96 m from the south edge of the intersection when it stops. (b) The blue car needs to travel (28 + 4.52) m = 32.52 m from the south edge of the intersection in oder to completely clear from that intersection. Use equation: s = ut + (1/2)at^2 with s = 32.52 m, u = 12.29 m/s, a = -2.1 m/s^2, t = ? hence, 32.52 = 12.29t + (1/2).(-2.1)t^2 solve for t gives t = 7.663 s or -4.04 s (rejected) Therefore, the blue car needs 7.663 s to clear off the intersection. (c) Consider the red car. Use equation: s = ut + (1/2)at^2 with u = 0 m/s, t = 7.663 s, a = 5.6 m/s^2, s = ? hence, s = (1/2).(5.6).(7.663^2) m = 164.4 m The red car is at 164.4 m from the west edge of the intersection. (d) Use equation: v = u + at with u = 0 m/s, a = 5.6 m/s^2, t = 7.663 s, v =? hence, v = 5.6 x 7.663 m/s = 42.9 m/s The speed of the red car is 42.9 m/s
其他解答:
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