A few quick problems with kinematics

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The discussion revolves around resolving kinematics problems where the original poster expresses confusion over their calculations. For the truck's travel distance, the correct answer is 1200 m, which requires accounting for the constant speed segment. The height of the building is correctly calculated as 131 m, while the average velocity of the bus is 50 km/hr, factoring in the stop time. The time at which a particle is at rest is confirmed to be 3 seconds, and the average acceleration over a specified interval is –18 m/s², emphasizing the need to differentiate between average and instantaneous values. Clarifications on these problems highlight common pitfalls in applying kinematic equations.
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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.
 
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