Answer:
0.5m
Explanation:
Given,
Initial velocity u= 9m/s
Acceleration a=2m/s²
Using formula—
[tex]sn = u + \frac{1}{2} (2n - 1)a[/tex]
The particle displacement in 5 th sec is —
[tex]s5 = 9 + 0.5(10 - 1)( - 2)[/tex]
[here we have written ½ in the form of decimal,,0.5]
[tex] = 0[/tex]
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 formula —
[tex]v = u + at[/tex]
Inserting values---
⇒
[tex]0 = 9 - 2 \times t[/tex]
⇒
[tex]0 = 9 - 2t[/tex]
⇒
[tex]2t = 9 - 0[/tex]
[changing the sign of 2t as it is placed before the =]
⇒
[tex]2t = 9[/tex]
⇒
[tex]t = \frac{9}{2} [/tex]
⇒
[tex]t = 4.5 \: seconds[/tex]
The velocity at 4 sec is ----
⇒
[tex]v = u + at[/tex]
[inserting values---]
⇒
[tex]v = 9 + ( - 2) \times 4[/tex]
⇒
[tex]v = 9 + ( - 8)[/tex]
⇒
[tex]v = 1 \: m {s}^{ - 1} [/tex]
Using formula,
[tex]s = ut + \frac{1}{2} a {t}^{2} [/tex]
distance covered in time 4 sec to 4.5 sec is
[tex]s1 = 1 \times \frac{1}{2} + \frac{1}{2} \times ( - 2) \times { (\frac{1}{2}) }^{2} [/tex]
[tex] = 0.25m[/tex]
And distance covered in time 4.5 sec to 5 sec is
⇒
[tex]s2 = \frac{1}{2} \times 2 \times { (\frac{1}{2}) }^{2} [/tex]
[tex] = 0.25m[/tex]
[as velocity at 4.5 is zero]
Thus, total distance covered in 5 th sec
[tex] = s1 + s2[/tex]
[tex] = 0.25 + 0.25[/tex]
[tex] = 0.5[/tex]
Hence, the distance covered in the 5th second of it's motion is 0.5m
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