L2 Displacement, Velocity, and Acceleration - Part 8
Completion requirements
Unit 7B
Integrals Part 2
Lesson 2: Displacement, Velocity, and Acceleration
An object is moving in a straight line. Its position, relative to a fixed point, is given by «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«msup»«mi»t«/mi»«mn»3«/mn»«/msup»«mo»§#8722;«/mo»«mn»4«/mn»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»12«/mn»«mi»t«/mi»«/math», where «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/math» is in metres and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math» is in seconds. Describe the motion when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»1«/mn»«/math».
First, find the velocity and acceleration functions at any time «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math».
Now, find the displacement, velocity, and acceleration at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»1«/mn»«/math».
At «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»1«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math», the object’s position is at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»9«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math», its velocity is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»7«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math», and its acceleration is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»§#8722;«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/math».
The object can be found «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»9«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» to the right of the fixed point. Since the velocity is positive, the object is moving to the right. Because the acceleration is negative, and the object is moving in a positive direction, the object is slowing down. A negative acceleration doesn’t always mean that an object is slowing down. It does in this situation, but this is because it is moving in a positive direction.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mi»t«/mi»«mn»3«/mn»«/msup»«mo»§#8722;«/mo»«mn»4«/mn»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»12«/mn»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»§#8722;«/mo»«mn»8«/mn»«mi»t«/mi»«mo»+«/mo»«mn»12«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»a«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»6«/mn»«mi»t«/mi»«mo»§#8722;«/mo»«mn»8«/mn»«/mtd»«/mtr»«/mtable»«/math»
Now, find the displacement, velocity, and acceleration at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»1«/mn»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mi»t«/mi»«mn»3«/mn»«/msup»«mo»§#8722;«/mo»«mn»4«/mn»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»12«/mn»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»s«/mi»«mfenced»«mn»1«/mn»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mfenced»«mn»1«/mn»«/mfenced»«mn»3«/mn»«/msup»«mo»§#8722;«/mo»«mn»4«/mn»«msup»«mfenced»«mn»1«/mn»«/mfenced»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»12«/mn»«mfenced»«mn»1«/mn»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd/»«mtd/»«/mtr»«mtr»«mtd»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»§#8722;«/mo»«mn»8«/mn»«mi»t«/mi»«mo»+«/mo»«mn»12«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«mfenced»«mn»1«/mn»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«msup»«mfenced»«mn»1«/mn»«/mfenced»«mn»2«/mn»«/msup»«mo»§#8722;«/mo»«mn»8«/mn»«mfenced»«mn»1«/mn»«/mfenced»«mo»+«/mo»«mn»12«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»7«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd/»«mtd/»«/mtr»«mtr»«mtd»«mi»a«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»6«/mn»«mi»t«/mi»«mo»§#8722;«/mo»«mn»8«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»a«/mi»«mfenced»«mn»1«/mn»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»6«/mn»«mfenced»«mn»1«/mn»«/mfenced»«mo»§#8722;«/mo»«mn»8«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«mn»2«/mn»«/mtd»«/mtr»«/mtable»«/math»
At «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»1«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math», the object’s position is at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»9«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math», its velocity is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»7«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math», and its acceleration is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»§#8722;«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/math».
The object can be found «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»9«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» to the right of the fixed point. Since the velocity is positive, the object is moving to the right. Because the acceleration is negative, and the object is moving in a positive direction, the object is slowing down. A negative acceleration doesn’t always mean that an object is slowing down. It does in this situation, but this is because it is moving in a positive direction.