Physics Solutions

Solutions 

1- The three equations are,

1. v = u + at
2. v² = u² + 2as
3. s = ut + ½at²

Derivation of the Equations of Motion

1. v = u + at

Let us begin with the first equation, v=u+at. This equation only talks about the acceleration, time, the initial and the final velocity. Let us assume a body that has a mass “m” and initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. Now we know that:

Acceleration = Change in velocity/Time Taken

Therefore,  Acceleration = (Final Velocity-Initial Velocity) / Time Taken

Hence, a = v-u /t or at = v-u

Therefore, we have: v = u + at

2. v² = u² + 2as

We have, v = u + at. Hence, we can write t = (v-u)/a

Also, we know that, Distance = average velocity × Time

Therefore, for constant acceleration we can write: 

Average velocity = (final velocity + initial velocity)/2 = (v+u)/2

Hence, Distance (s) = [(v+u)/2]  × [(v-u)/a]

or  s = (v² – u²)/2a

or 2as = v² – u²

or v² = u² + 2as
3. s = ut + ½at²

Let the distance be “s”. We know that

Distance = Average velocity × Time. Also, Average velocity = (u+v)/2

Therefore, Distance (s) = (u+v)/2 × t

Also, from v = u + at, we have:

s = (u+u+at)/2 × t = (2u+at)/2 × t

s = (2ut+at²)/2 = 2ut/2 + at²/2

or s = ut +½ at²

Solution 8 = > as we know a=v-u/t
∴ 8–20/2 = -6 = a
⇒ the velocities in 1sec time interval form AP.
∴ in 1st sec, d=44m ,2nd sec v is 38m…….
again,
we get AP till the 6th second.
∴ total distance = 44+38+32+26+20+14+8=182m….

Solution 8 = For instance :- take downward side as positive.

s = +100m

g = +10m/s²

u = 0

v = ?

Apply v²-u²= 2as

Plug-in the values , we get :-

v²-0 = 2×10×100

v² = 2000

=> v = ±√2000 m/s

v can't be negative as motion is in positive direction!

Hence v = √2000 m/s