Derive the formula S-ut +1/2, at^2 where the symbols have usual meanin

1.	Derive the formula S-ut +1/2, at^2 where the symbols have usual meanings.A train staring from stationary position and moving with uniform accelerationattains a speed of 36 km per hour in 10 minutes. Find its acceleration.
Answer» ☆ ΛПƧЩΣЯ ☆ Consider an object moving with initial velocity u, is uniformly ACCELERATED with an acceleration 'a' for time 't', finally REACHES velocity 'v' after covering displacement 's'. From FIGURE OA=OC=u. BC=v OC=AD=t. BD=v-u The displacement of an object can be calculated by finding the area under velocity time graph. Area under velocity time graph =Area of trapezium OABC ✈ Displacement =Area of rectangle OA DC+Area of triangle ABCD. Displacement =length +breadth+1/2×base×height S=OA ×OC+1/2 AD×BD =ut + 1/2 t (v-u) =ut + 1/2 t ×a t $s = u \: t + \frac{1}{2} a {t}^{2}$ ✈ 5 thanks ✈️ ➡️ Fo I l o W me ⬅️

Derive the formula S-ut +1/2, at^2 where the symbols have usual meanings.A train staring from stationary position and moving with uniform accelerationattains a speed of 36 km per hour in 10 minutes. Find its acceleration.

Answer»

☆ ΛПƧЩΣЯ ☆

Consider an object moving with initial velocity u, is uniformly ACCELERATED with an acceleration 'a' for time 't', finally REACHES velocity 'v' after covering displacement 's'.

From FIGURE OA=OC=u. BC=v

OC=AD=t. BD=v-u

The displacement of an object can be calculated by finding the area under velocity time graph.

Area under velocity time graph =Area of trapezium OABC

✈ Displacement =Area of rectangle OA DC+Area of triangle ABCD.

Displacement =length +breadth+1/2×base×height

S=OA ×OC+1/2 AD×BD

=ut + 1/2 t (v-u)

=ut + 1/2 t ×a t

$s = u \: t + \frac{1}{2} a {t}^{2}$

✈ 5 thanks ✈️

➡️ Fo I l o W me ⬅️

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