in our day to day life we experience or some across different type of motions can you give an example of the motion when the acceleration is uniform or the rate of acceleration is constant
Answers
Compared to displacement and velocity, acceleration is like the angry, fire-breathing dragon of motion variables. It can be violent; some people are scared of it; and if it's big, it forces you to take notice. That feeling you get when you're sitting in a plane during take-off, or slamming on the brakes in a car, or turning a corner at a high speed in a go kart are all situations where you are accelerating.
Acceleration is the name we give to any process where the velocity changes. Since velocity is a speed and a direction, there are only two ways for you to accelerate: change your speed or change your direction—or change both.
If you’re not changing your speed and you’re not changing your direction, then you simply cannot be accelerating—no matter how fast you’re going. So, a jet moving with a constant velocity at 800 miles per hour along a straight line has zero acceleration, even though the jet is moving really fast, since the velocity isn’t changing. When the jet lands and quickly comes to a stop, it will have acceleration since it’s slowing down. [Wait, what?]
Or, you can think about it this way. In a car you could accelerate by hitting the gas or the brakes, either of which would cause a change in speed. But you could also use the steering wheel to turn, which would change your direction of motion. Any of these would be considered an acceleration since they change velocity. [Huh?]
What's the formula for acceleration?
To be specific, acceleration is defined to be the rate of change of the velocity.
\Huge{a=\frac{\Delta v}{\Delta t} = \frac {v_f-v_i}{\Delta t}}a=
Δt
Δv
=
Δt
v
f
−v
i
a, equals, start fraction, delta, v, divided by, delta, t, end fraction, equals, start fraction, v, start subscript, f, end subscript, minus, v, start subscript, i, end subscript, divided by, delta, t, end fraction
The above equation says that the acceleration, aaa, is equal to the difference between the initial and final velocities, v_f - v_iv
f
−v
i
v, start subscript, f, end subscript, minus, v, start subscript, i, end subscript, divided by the time, \Delta tΔtdelta, t, it took for the velocity to change from v_iv
i
v, start subscript, i, end subscript to v_fv
f
v, start subscript, f, end subscript. [Really?]
Note that the units for acceleration are \dfrac{\text m/s}{\text s}
s
m/s
start fraction, m, slash, s, divided by, s, end fraction , which can also be written as \dfrac{\text m}{\text s^2}
s
2
m
start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction. That's because acceleration is telling you the number of meters per second by which the velocity is changing, during every second. Keep in mind that if you solve \Large{a= \frac {v_f-v_i}{\Delta t}}a=
Δt
v
f
−v
i
a, equals, start fraction, v, start subscript, f, end subscript, minus, v, start subscript, i, end subscript, divided by, delta, t, end fraction for v_fv
f
v, start subscript, f, end subscript, you get a rearranged version of this formula that’s really useful.
v_f=v_i+a\Delta tv
f
=v
i
+aΔtv, start subscript, f, end subscript, equals, v, start subscript, i, end subscript, plus, a, delta, t
This rearranged version of the formula lets you find the final velocity, v_fv
f
v, start subscript, f, end subscript, after a time, \Delta tΔtdelta, t, of constant acceleration, aaa.