A particle has initial velocity 9 m/s due east and constant acceleration 2m/s^2 due west. if the distance covered by it in the 5th second of its motion is n/10 m, then find the value of 'n'
Answers
Answer:
Further explanation: The particle is moving to east (right) with velocity 9m/s and a constant acceleration of 2m/s^2 due west which means deceleration with 2m/s^2. Therefore a=-2.
Explanation:
Given, initial velocity,
u
=9
i
^
m/s and acceleration,
a
=−2
i
^
m/s
2
Using formula, S
n
=u+
2
1
(2n−1)a, the particle displacement in 5 th sec is
S
5
=9
i
^
+0.5(10−1)(−2
i
^
)=0
Thus, the particle will be reached in the initial position in 5 th sec.
Let, the velocity will be zero at time t.
So using, v=u+at⇒0=9−2t or t=4.5sec
The velocity at 4 sec is v=9−2(4)=1m/s
Using formula, S=ut+(1/2)at
2
, distance covered in time 4 sec to 4.5 sec is S
1
=1(1/2)+(1/2)(−2)(1/2)
2
=0.25m and distance covered in time 4.5 sec to 5 sec is S
2
=(1/2)(2)(1/2)
2
=0.25m (as velocity at 4.5 sec is zero)
Thus, total distance covered in 5 th sec S
1
+S
2
=0.25+0.25=0.5m
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