The correct relationship between the displacement, s, of a particle initially at rest in a linear motion and the time, t, is?
a
s \(\alpha\) t\(^{-2}\)
b
s \(\alpha\) t\(^{-1}\)
c
s \(\alpha\) t
d
s \(\alpha\) t\(^{2}\)
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A1_is_the_goal
3 years ago
Since the particle is in linear motion, that is, linear acceleration, a=v/t---(1)
Where a= acceleration, v= velocity and t= time.
But velocity = displacement /time
v=s/t---(2)
Inputing the value of v from equ.(2) into equ.(1)...
a=v/t
a= (s/t)/t
a=s/t²
Assuming that a is constant...
s/t²=a/1
Cross multiplying
s=at²
Remember that a is constant
Therefore: s α t²
{s is directly proportional to t²}
I hope this helped you understand...
