Q18.A particle is moving along a straight line an
its position is given by the relation x =
15t +40)m. Find (a) The time at which velocity is
zero.
(b) Position and displacement of the particle at
that point. time
Tire
(c) Acceleration for the particle at that line.
Answers
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1
Answer:
t = x^1/2 + 3
√ x = t - 3
sq both sides
x = ( t - 3 )^2
x = ______(1)
we know
differentiation of distance w.r.t time ,give velocity
so
d (x) / dt = 2 t - 6
v = 2 t - 6
v = 0 ,given
t = 3 second
from eq 1
x = t^2 - 6t + 9
x = 3 × 3 - 6 × 3 + 9
x = 9 - 18 + 9
x = 0
SO DISPLACEMENT WILL BE ZERO
HOPE HELPED
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Explanation:
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