Question

Q123.A body travelling at 15 ms 1 accelerates and attains a speed of 27 m s -1 in 5 s. The distance travelled by the body in 5th second is.
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Answer 1

Answer:

25.8 m

Explanation:

$u = 15 ms^{-1}$

$v = 27 ms^{-1}$

$t = 5 s$

$v = u + at$

$27 = 15 + a * 5$

$12 = 5a$

$a = \frac{12}{5}$

a = $2.4 ms^{2}$

$S_{n} = u +\frac{a}{2} (2n-1)$

$S_{5} = 15 + \frac{2.4}{2} (2*5 - 1)$

$S_{5} = 15 + 1.2 (10-1)$

$S_{5} = 15 + 1.2*9$

$S_{5} = 15+10.8$

$S_{5} = 25.8$

Answer 2

Answer:

25.8 m

Explanation:

Since body is uniformly accelerated

so, acceleration = (final velocity - initial velocity) / time taken

acceleration = (27 - 15) / 5 = 2.4 $ms^{-2}$

Distance travelled in  nth second = $u + \frac{a}{2} (2n - 1)$

here, u = 15 m/s, a = 2.4 $ms^{-2}$ , n = 5

So, distance travelled in 5th second = $15 + \frac{2.4}{2} X (2X5 - 1)$

                      = 15 + 1.2 X 9 = 15 + 10.8 = 25.8 m

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