A few quick problems with kinematics

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SUMMARY

The forum discussion addresses common problems in kinematics, specifically focusing on calculations involving acceleration, velocity, and position. Key problems include determining the distance traveled by a truck under acceleration and deceleration, the height of a building based on the speed of a falling ball, and average velocity calculations for a bus trip. The correct answers for the discussed problems are 1200 m for the truck, 131 m for the building, and 50 km/hr for the bus's average velocity. Misunderstandings regarding the inclusion of constant speed segments and the distinction between average and instantaneous acceleration are also clarified.

PREREQUISITES
  • Understanding of kinematic equations, including x = v0t + 1/2at²
  • Knowledge of average velocity calculations using v = (delta)x / (delta)t
  • Familiarity with concepts of acceleration as the second derivative of position
  • Ability to apply gravitational acceleration (–9.8 m/s²) in free-fall problems
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  • Study the derivation and application of kinematic equations in various scenarios
  • Learn how to calculate average and instantaneous acceleration from position functions
  • Explore real-world applications of kinematics in vehicle motion analysis
  • Investigate the effects of air resistance on falling objects and their impact on calculations
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Students and educators in physics, particularly those focusing on mechanics, as well as anyone preparing for exams involving kinematics and motion analysis.

m00nbeam360
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Hi there,

Sorry for so many questions, but I feel frankly stupid for getting these wrong, and yet I have no clue why they're so off. Any ideas of why the answers are what they are? Thanks so much.
15. A truck starts from rest and accelerates at 1.5 m/s2for 20 s. It then travels at constant speed for 20 s and decelerates at –1.5 m/s2until it stops. How far did the truck travel?

a. 300 m

b. 600 m

*c. 1200 m

d. 1500 m


I got 600 because I used x = 1/2at^2 + v0t + x0 with v0 = 0, a = 1.5 m/s^2, t = 20 s, and x = ?

17. A ball is dropped from the top of a tall building. Just before it strikes the ground it is moving at 34.3 m/s. How tall is the building?

a. 60 m

*b. 131 m

c. 250 m

d. 125 m


I got 60 m using v^2 = 2ax with v0 = 0, vf = 34.3 m/s, a = -9.8 m/s^2, and x = ?

18. A bus leaves a terminal en-route to a destination 200 km away. During the trip, the bus stopped for a 45 minute lunch. The bus arrived at its destination four hours after it departed. The average velocity was:

a. 0 km/hr

b. 61 km/hr

c. 30 km/hr

*d. 50 km/hr


I used v = (delta)x / (delta)t with delta x = 200 km, and delta t = 3.25 hr

19. The position of a particle is given by x ( t ) =24 t – 4 t 2. At what time is the particle at rest?

*a. 3s

b. 8s

c. 10s

d. 2s


I got 6, Wolfram got 6, I'm very confused.

20. The position of a particle as a function of time is given by 4 t +3 t 2–2 t 3. What is the average acceleration over the time interval from 0 to 4s?

a. 90 m/s2

*b. –18m/s2

c. –102 m/s2

d. 102 m/s2


Since acceleration is the second-derivative of position, I got 6 - 12t. Which apparently is wrong.

Thanks again.
 
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m00nbeam360 said:

15. A truck starts from rest and accelerates at 1.5 m/s2for 20 s. It then travels at constant speed for 20 s and decelerates at –1.5 m/s2until it stops. How far did the truck travel?

a. 300 m

b. 600 m

*c. 1200 m

d. 1500 m


I got 600 because I used x = 1/2at^2 + v0t + x0 with v0 = 0, a = 1.5 m/s^2, t = 20 s, and x = ?
Did you include the constant speed segment? (I'd say that none of the answers are correct.)

17. A ball is dropped from the top of a tall building. Just before it strikes the ground it is moving at 34.3 m/s. How tall is the building?

a. 60 m

*b. 131 m

c. 250 m

d. 125 m


I got 60 m using v^2 = 2ax with v0 = 0, vf = 34.3 m/s, a = -9.8 m/s^2, and x = ?
Your answer is correct.

18. A bus leaves a terminal en-route to a destination 200 km away. During the trip, the bus stopped for a 45 minute lunch. The bus arrived at its destination four hours after it departed. The average velocity was:

a. 0 km/hr

b. 61 km/hr

c. 30 km/hr

*d. 50 km/hr


I used v = (delta)x / (delta)t with delta x = 200 km, and delta t = 3.25 hr
The zero-speed segment counts. Δt = 4 hours.
19. The position of a particle is given by x ( t ) =24 t – 4 t 2. At what time is the particle at rest?

*a. 3s

b. 8s

c. 10s

d. 2s


I got 6, Wolfram got 6, I'm very confused.
The speed is zero, not the position.

20. The position of a particle as a function of time is given by 4 t +3 t 2–2 t 3. What is the average acceleration over the time interval from 0 to 4s?

a. 90 m/s2

*b. –18m/s2

c. –102 m/s2

d. 102 m/s2


Since acceleration is the second-derivative of position, I got 6 - 12t. Which apparently is wrong.
They want the average acceleration, not the instantaneous acceleration.
 
http://img15.imageshack.us/img15/2332/p1010016rn.jpg
 
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