Solution: By using Newton's First Law of Motion: VFinal = Uinitial + (acceleration x time) Part 1: Let us integrate that the car with an initial velocity of 13.89 m/s comes to a stop at 3 seconds and reaches a final velocity of 0 m/s2 This event is calculated as follows: 0m/s2 = 13.89 m/s + (a x 3s) %3D a = S-13.89 dt -13.89t +c a = -13.89 () + (0 m/s2) [c, = 0 m/s: a = -13.89 t+0 m/s²] 13.89 a = - 3 1 - t= 3 a = -4. 63 m/s2 The distance traveled during this time is given by: s = Vinitial X time + [5 (acceleration) x (time) ] s = (-13.89 t+ 0m/s ) dt (-13.89 t + 0 m/s*) dt --13.89t + s = (13.89 m/s) x (3 s) +[(-4.63 m/s) x (3 s) ] s = 41.67 m - 20.835 m S 20.84 m
Solution: By using Newton's First Law of Motion: VFinal = Uinitial + (acceleration x time) Part 1: Let us integrate that the car with an initial velocity of 13.89 m/s comes to a stop at 3 seconds and reaches a final velocity of 0 m/s2 This event is calculated as follows: 0m/s2 = 13.89 m/s + (a x 3s) %3D a = S-13.89 dt -13.89t +c a = -13.89 () + (0 m/s2) [c, = 0 m/s: a = -13.89 t+0 m/s²] 13.89 a = - 3 1 - t= 3 a = -4. 63 m/s2 The distance traveled during this time is given by: s = Vinitial X time + [5 (acceleration) x (time) ] s = (-13.89 t+ 0m/s ) dt (-13.89 t + 0 m/s*) dt --13.89t + s = (13.89 m/s) x (3 s) +[(-4.63 m/s) x (3 s) ] s = 41.67 m - 20.835 m S 20.84 m
College Physics
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Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Given the initial velocity as 13.89 m/s, can you perform/integrate the solution/function of acceleration and distance based on the typewritten formula / highlighted parts. Thank you
![Solution:
By using Newton's First Law of Motion:
Final =Uinitial + (acceleration x time
Part 1: Let us integrate that the car with an initial velocity of 13.89 m/s comes to a stop
at 3 seconds and reaches a final velocity of 0 m/s2
This event is calculated as follows:
0m/s² = 13.89 m/s + (a x 3s)
a = J-13.89 dt -13.89t +c,
a = -13.89 ()+ (0 m/s)
- [c, = 0 m/s*: a = -13.89 t+0 m/s²]
13.89
1
3
3.
a = -4.63 m/s2
The distance traveled during this time is given by:
s = Vinirial X time + [,(acceleration) x (time) ]
2.
s = (-13.89 t+0 m/s ) dt (-
-13.89 t+0 m/s) dt -13.89t +
%3D
s = (13.89 m/s) x (3 s) +[(-4.63 m/s) x (3 s) ]
s= 41.67 m- 20.835 m
s 20.84 m
Le](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fea40c60e-8df4-4eba-ba3b-17ca1c259944%2F3cba5124-0dd8-47c7-a494-a3a97f2abe9d%2Fsppyqce_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Solution:
By using Newton's First Law of Motion:
Final =Uinitial + (acceleration x time
Part 1: Let us integrate that the car with an initial velocity of 13.89 m/s comes to a stop
at 3 seconds and reaches a final velocity of 0 m/s2
This event is calculated as follows:
0m/s² = 13.89 m/s + (a x 3s)
a = J-13.89 dt -13.89t +c,
a = -13.89 ()+ (0 m/s)
- [c, = 0 m/s*: a = -13.89 t+0 m/s²]
13.89
1
3
3.
a = -4.63 m/s2
The distance traveled during this time is given by:
s = Vinirial X time + [,(acceleration) x (time) ]
2.
s = (-13.89 t+0 m/s ) dt (-
-13.89 t+0 m/s) dt -13.89t +
%3D
s = (13.89 m/s) x (3 s) +[(-4.63 m/s) x (3 s) ]
s= 41.67 m- 20.835 m
s 20.84 m
Le
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