Answer:
Displacement of Mr. Llama: Option D. 0 miles.
Explanation:
The magnitude of the displacement of an object is equal to the distance between its final position and its initial position. In other words, as long as the initial and final positions of the object stay unchanged, the path that this object took will not affect its displacement.
For Mr. Llama:
- Final position: Mr. Llama's house;
- Initial position: Mr. Llama's house.
The distance between the final and initial position of Mr. Llama is equal to zero. As a result, the magnitude of Mr. Llama's displacement in the entire process will also be equal to zero.
We can’t see the following
Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)

where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:

0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:

Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:

5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:

10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:

10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]
Answer:
5.3 m/s
Explanation:
First, find the time it takes for him to fall 7m.
y = y₀ + v₀ t + ½ at²
0 = 7 + (0) t + ½ (-9.8) t²
0 = 7 − 4.9 t²
t ≈ 1.20 s
Now find the velocity he needs to travel 6.3m in that time.
x = x₀ + v₀ t + ½ at²
6.3 = 0 + v₀ (1.20) + ½ (0) (1.20)²
v₀ ≈ 5.27 m/s
Rounded to two significant figures, the man must run with a speed of 5.3 m/s.