The rate at which velocity changes with time; the change in velocity may be in magnitude or direction or both is called acceleration motion. QUESTION 18 What is the acceleration of a car that maintains a constant velocity of 100.0 km/s for 10.0 s? A. zero km/s^2 B. 100.0 km/hr.s C. 10.0 km/hr D. 10.0 km/hr.s QUESTION 19 What is the speed acquired by a freely falling object after being dropped from a rest position (m)? A. 10.0 m/s B. 50.0 m/s C. 58.8 m/s D. 9.8 m/s
The rate at which velocity changes with time; the change in velocity may be in magnitude or direction or both is called acceleration motion. QUESTION 18 What is the acceleration of a car that maintains a constant velocity of 100.0 km/s for 10.0 s? A. zero km/s^2 B. 100.0 km/hr.s C. 10.0 km/hr D. 10.0 km/hr.s QUESTION 19 What is the speed acquired by a freely falling object after being dropped from a rest position (m)? A. 10.0 m/s B. 50.0 m/s C. 58.8 m/s D. 9.8 m/s
The Correct Answer and Explanation is:
Let’s tackle the two questions one by one:
Question 18: What is the acceleration of a car that maintains a constant velocity of 100.0 km/s for 10.0 s?
When a car moves with a constant velocity, its acceleration is zero. Acceleration is the rate at which velocity changes, so if the velocity is constant (i.e., it doesn’t change in magnitude or direction), there is no change in velocity, and hence no acceleration.
Correct Answer: A. zero km/s²
Explanation:
Acceleration is defined as the rate of change of velocity. The formula for acceleration is:a=ΔvΔta = \frac{{\Delta v}}{{\Delta t}}a=ΔtΔv
Where:
- Δv\Delta vΔv is the change in velocity,
- Δt\Delta tΔt is the change in time.
In this case, since the velocity is constant (100.0 km/s), Δv=0\Delta v = 0Δv=0. So, no matter how much time passes, the velocity doesn’t change. Therefore, the acceleration is zero.
Question 19: What is the speed acquired by a freely falling object after being dropped from a rest position?
For a freely falling object, the speed it acquires depends on the time it has been falling, and it increases due to gravitational acceleration. The acceleration due to gravity is approximately 9.8 m/s29.8 \, \text{m/s}^29.8m/s2 on Earth.
To calculate the speed after a certain amount of time, we use the formula:v=g×tv = g \times tv=g×t
Where:
- vvv is the final velocity (speed),
- ggg is the acceleration due to gravity (9.8 m/s29.8 \, \text{m/s}^29.8m/s2),
- ttt is the time.
If the object has fallen for 1 second, the speed would be:v=9.8 m/s2×1 s=9.8 m/sv = 9.8 \, \text{m/s}^2 \times 1 \, \text{s} = 9.8 \, \text{m/s}v=9.8m/s2×1s=9.8m/s
For longer times, the speed increases proportionally.
Correct Answer: D. 9.8 m/s
Explanation:
When an object is dropped from a height, it accelerates due to gravity. In the absence of air resistance, the object starts from rest and speeds up as it falls, with the speed increasing by 9.8 m/s9.8 \, \text{m/s}9.8m/s each second. Therefore, after 1 second, the speed is 9.8 m/s. For the question provided, the speed after being dropped from a rest position would be approximately 9.8 m/s.
