Question

A particle moving with uniform acceleration has a displacement of (3.8+0.4n) m in the nth second of its travel .Then its velocity at the end of 5th second is m/s is:_
a.5.8
b.6
c.4.8
d.8

Answer 1

Answer:

(b) 6 m/s

Explanation:

So there's a formula for such questions :

$S_{nth} = u + \frac{1}{2} a(2n -1)$ ; also know as Fourth equation of Motion

$S_{nth}$ =  (3.8 + 0.4n) as per the question,

⇒ (3.8 + 0.4n) =  $u + \frac{1}{2} a(2n -1)$

RHS :

= $u + an- \frac{a}{2}$

Rearrange according to LHS -

= $(u - \frac{a}{2})+ an$

Compare with LHS :

⇒ (3.8 + 0.4n) = $(u - \frac{a}{2})+ an$

⇒ 3.8 = $(u - \frac{a}{2})$ --------(1)

and 0.4n = $an$  ⇒ 0.4 = $a$ ------(2)

Now we know that acceleration (a) = 0.4 $m/s^{2}$

Put the value of $a$ in equation (1)

⇒ 3.8 = $u - \frac{(0.4)}{2}$

⇒ 3.8 = $u - 0.2$

⇒ $u$ = 4

Now we know that :

Initial velocity (u) = 4 m/s

and Acceleration (a) = 0.4 $m/s^{2}$

But we need to find the veocity at 5th second or velocity at t = 5s. Let's use 1st equation of motion to find that:

⇒ v = u +at

⇒ v = 4 + 0.4 (5)

⇒ v = 4 + 2

⇒ v = 6 m/s

Therefore, velocity at the end of 5th second is 6 m/s. Which is option (b).

Hope it helps :)

(If you find any mistakes, which I am sure there aren't any, then please contact me within 10 minutes of writing this answer, if not possible then please forgive me).

If you have any other queries, feel free to disturb me again.