Question

derive the equation ignition V=Vo+at from v-tgraph​

Answer 1

GIVEN :-

• v = u + at

TO DERIVE :-

• First equation of motion (v = u + at)

DERIVATION :-

$\setlength{\unitlength}{2mm}\begin{picture}(6,4)\linethickness{0.26mm} \put(1,7){\dashbox{0.07}(18,17)} \put(1,1.02){\dashbox{0.07}(18,6)}\put(1,1){\vector(2,0){30}}\put(1,1){\vector(0,0){25}}\qbezier(1,7)(1,7)(19,24)\qbezier(4,7)(6,7)(4,10)\put(5.5,8){ \sf \theta }\put(18.5, - 0.8){ \sf C }\put(19.5,6){ \sf D }\put(18.5,25){ \sf B }\put(1.2,22){ \sf E }\put(1.4,25){ \sf y }\put( - 0.2,6){ \sf A }\put( - 0.2, - 0.4){ \sf O }\put(31.2, 0.5){ \sf x }\put(22,17){\vector(0,0){8}}\put(22,15){\vector(0, - 1){8}}\put(20,15.5){ \sf (v - u) }\put(28,9){\vector(0,0){15}}\put(28,6){\vector(0, - 1){5}}\put(27,7){ \sf (V) }\put( - 7,15){ \textsf{ \textbf{Velocity}} }\put( 8, - 0.5){ \textsf{\textbf{Time}} }\end{picture}$

Let us consider an object is moving with uniform acceleration along a straight line . Let (u) be the initial velocity at t = 0 . After interval of time (t) , its velocity becomes u.

➳ Let OA = u [ t = 0 ]

Here in the figure we draw AD ⟂ BC , and BE ⟂ OY.

➳ ∠BAD = ϴ. [ say ]

Now,

➳ Acceleration = slope of Line AB

➳ a = tan ϴ

➳ a = BD/AD. [ tan ϴ = P/B ]

➳ a = (v - u)/t. [ AD = OC = t ]

or, at = v - u.

or, v = u + at. [ 1st Equation of motion ]

Hence we proved the required equation of motion (v = u + at)

Answer 2

कंसीडर ए पार्टिकल स्टार्टिंग फ्रॉम और उज्जैन जीरो हैविंग इस सेम इंडिविजुअल वेलोसिटी वी जीरो वेलोसिटी इज इक्वल टू जीरो सेकेंड लेट ए पार्टिकल B5 जेनरेटेड यूनीफामर्ली बाय हेविंग फाइनल वेलोसिटी वी लेटी सेकंड फ्रॉम दिग्राफ कंसीडर एबीसी