If the acceleration is constant (negative or positive) the instantaneous acceleration cannot be
Average acceleration: [final velocity - initial velocity ] /Δ time
Instantaneous acceleration = d V / dt =slope of the velocity vs t graph
If acceleration is increasing, the slope of the curve at one moment will be higher than the average acceleration.
If acceleration is decreasing, the slope of the curve at one moment will be lower than the average acceleration.
If acceleration is constant, the acceleration at any moment is the same, then only at constant accelerations, the instantaneuos acceleration is the same than the average acceleration.
Constant zero acceleration is a particular case of constant acceleration, so at constant zero acceleration the instantaneous accelerations is the same than the average acceleration: zero. But, it is not true that only at zero acceleration the instantaneous acceleration is equal than the average acceleration.
That is why the only true option and the answer is the option D. only at constant accelerations.
Answer:
V = 0.714m/s
Explanation:
Full solution calculation can be found in the attachment below.
From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.
Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.
See the attachment below for the solution calculation.
Answer:
= 5/9
Explanation:
This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ_liquid g V_liquid
let's write the translational equilibrium condition
B - W = 0
let's use the definition of density
ρ_body = m / V_body
m = ρ_body V_body
W = ρ_body V_body g
we substitute
ρ_liquid g V_liquid = ρ_body g V_body
In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar
V = A h_bogy
Thus
we substitute
5/9 = 
Explanation:
Distance covered by the particle is given by:
Distance (d) = rate (v) × time (t)
Speed of Mary, v₁ = 50 mph
Speed of Jim, v₂ = 60 mph
It is assumed that, Mary and Jim leave at the same time. After one hour, Jim is 10 miles ahead.
Distance travelled by Jim, d₁ = (60t + 10)
Distance travelled by Mary, d₂ = 50t
The distance between Mary and Jim is greater than or equal to 100 miles.



So, Jim takes is 9 hours more than Mary to cover same distance. Hence, this is the required solution.