Question

Establish the equation for position time relation(i.e, s=ut+1/2at^2)using velocity time graph. How is the
equation modified when
(i) a body is just dropped from some height, (ii) a body is thrown vertically upwards with some velocity ?​

Answer 1

Answer:

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Answer 2

Let a particle moves from point A with initial velocity u, motion of particle is accelerating { where acceleration is a}. so, after time t, velocity of particle becomes v = u + at. in this time particle reaches at point B.

now we have to find displacement covered by particle from point A to B.

we know,

displacement = area under the velocity-time graph

so, displacement covered by particle from point A to point B, s = area enclosed by velocity-time graph

= area of trapezium

= 1/2 ( sum of parallel sides) × distance between parallel sides.

= 1/2 × (u + u + at) × t

= 1/2 × (2u + at) × t

= ut + 1/2 at²

hence, s = ut + 1/2 at²

(i)a body is just dropping from some height

u = 0, a = -g

then, s = 0 × t + 1/2(-g)t²

s = -1/2gt² [ here negative sign indicates body moves downward]

(ii) a body is thrown upward with some velocity u,

here, a = -g

then, s = ut + 1/2(-g)t²

s = ut - 1/2gt²