A ) v = v o + a t ( the acceleration will be negative )
9.50 = 16.0 + a * 1.2
a * 1.2 = -16.0 + 9.50
a * 1.2 = - 6.5
a = - 6.5 : 1.2
a = - 5.4167 m/s²
F = m * a = 950 kg * 5.4167 m/s²
F = 5,145.8 N ( the average force exerted on a car during braking )
b ) d = v o - a t² / 2
d = 16.0 * 1.2 - ( 5.4167 * 1.2² / 2 ) =
= 19.20 - 3.90 = 15.30 m
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.
Speed is the distance traveled
time
but velocity is the change in distance
time
The air pressure inside the balloon increases as the number of particles increases.
