Question

A body is moving with uniform acceleration covers distance of 30m in 5th second and 80m in 8th second. calculate acceleration and initial velocity​

Answer 1

Answer:

Distance covered in t th second by a body is given by S = u +1/2a(2t -1 )

Just using this above formula

you may write as,

30= u+1/2a(10-1)....(1)

80= u+1/2a(16-1)....(2)

Substracting (1) from (2)

we get, 6a/2= 50

Or a = 50/3 m/s^2

putting the value of a in ....(1)

u = 30 - 25/3*9= -45 m/s

I hope,It was helpful

Answer 2

Given :

A body covers distance of 30m in 5th second.

A body covers distance of 80m in 8th second.

To find :

acceleration and initial velocity

Formula used:

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

where :-

u = initial velocity

a = acceleration

n = nth second

Solution :

A body covers distance of 30m in 5th second, so we put S5 = 30 and n = 5 in the formula :-

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

$30= u + \frac{1}{2} a((2 \times 5) - 1)$

$30= u + \frac{1}{2} a(10- 1)$

$30= u + \frac{9}{2} a$

$30 - \frac{9}{2} a= u ......(1)$

A body covers distance of 30m in 5th second, so we put S8 = 80 and n = 8 in the formula :-

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

$80= u + \frac{1}{2} a((2 \times 8) - 1)$

$80= u + \frac{1}{2} a(16 - 1)$

$80= u + \frac{15}{2} a$

$80 - \frac{15}{2} a= u ......(2)$

Now from ( 1 )and (2) :-

$30 - \frac{9}{2} a = 80 - \frac{15}{2} a$

$\frac{6}{2} a = 50$

$3 a = 50$

$a = \frac{50}{3} \: \frac{m}{ {s}^{2} }$

Now to find initial velocity put a = 50/3 in :-

$30 - \frac{9}{2} a= u$

$30 - (\frac{9}{2} \times \frac{50}{3}) = u$

$- 45 \: \frac{m}{s} = u$

ANSWER :-

acceleration = 50/3 m/s²

initial velocity= 45 m/s

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Learn more :-

v= u+at

v²-u²= 2as

s = ut+1/2at²

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