1.) What is the acceleration of a car that increases its velocity from 0 to +14m/s in 10s?
2.) What is the acceleration of a car that maintains a constant velocity of - 14m/s for 10s?
3.) Seagulls are often observed dropping clams and other shellfish from a height to the rocks below, as a mean of opening the shells. If a seagull drops a shell from rest at a height of 14.0m, how fast is the shell moving when it hits the rocks?
a = change in velocity/change in time
= (14 -0) / 10
2.) zero
3.) (1/2) m v^2 = m g h
so
v = sqrt (2 g h)
1.) To find the acceleration of the car, you can use the formula: acceleration = (final velocity - initial velocity) / time.
Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = +14 m/s
Time (t) = 10 s
Substituting the given values into the formula, we get:
acceleration = (14 m/s - 0 m/s) / 10 s
acceleration = 14 m/s / 10 s
acceleration = 1.4 m/s²
Therefore, the acceleration of the car is 1.4 m/s².
2.) In this case, the car maintains a constant velocity of -14 m/s. Since the velocity is constant, there is no acceleration. Acceleration is the rate of change of velocity, and if the velocity does not change, the acceleration is 0 m/s².
Therefore, the acceleration of the car in this case is 0 m/s².
3.) To find how fast the shell is moving when it hits the rocks, we can use the principle of conservation of energy. The potential energy at the top (when the shell is at rest) is converted into kinetic energy at the bottom (when the shell hits the rocks).
The potential energy (PE) of an object at height (h) is given by the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
Given:
Height (h) = 14.0 m
Let's assume the mass of the shell is m.
At the top, the potential energy is equal to the gravitational potential energy:
PE_top = m × g × h = m × 9.8 m/s² × 14.0 m
At the bottom, the potential energy is converted into kinetic energy:
KE_bottom = 0.5 × m × v²
Since the potential energy is converted entirely into kinetic energy, we can equate the two:
m × g × h = 0.5 × m × v²
Simplifying the equation, we get:
v² = 2gh
Substituting the values:
v² = 2 × 9.8 m/s² × 14.0 m
v² = 274.4 m²/s²
Taking the square root of both sides:
v ≈ 16.6 m/s
Therefore, the shell will be moving at approximately 16.6 m/s when it hits the rocks.