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My question: Using velocity-time graph derive the relation between position-time relation?
Answers
Answer:
Hey mate!
Here's your answer!!
Consider the graph shown in the given attachment. Obviously, the total distance s travelled by the object in time t is given by are under v-t graph.
Therefore, distance travelled, s = Area OABGC
Obviously, OABGC is a trapezium, whose area is (OA + BC/2) × OC.
But, OA = u, BC = v and OC = t
Distance travelled,
s = ( \frac{oa + bc}{2} ) \times oc = ( \frac{u \times v}{2} ) \times ts=(
2
oa+bc
)×oc=(
2
u×v
)×t
From the velocity - time relation, we have
at = v - u or t = v - u/a
On substituting this value of t in equation, we get
s = ( \frac{u + v}{2} ) \times ( \frac{v - u}{a} ) = \frac{v {}^{2} - u {}^{2} }{2a}s=(
2
u+v
)×(
a
v−u
)=
2a
v
2
−u
2
]
➡ v² - u² = 2as
Therefore, this equation is the position velocity relation for uniformly accelerated motion.
Hope it helps you!
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