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

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11th

Physics

Motion in a Straight Line

Distance and Displacement

A body moving with a unifor...

PHYSICS

A body moving with a uniform acceleration describes 12 m in the third second of its motion and 20 m in the fifth second. Find the distance covered in the length second.

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ANSWER

displacement for the n

th

second is given by ,

S

n

=u+

2

1

a(2n−1)

so from the question we get ,

S

3

=u+

2

1

a(2×3−1)⇒12=u+(5/2)a....(1)

S

5

=u+

2

1

a(2×5−1)⇒20=u+(9/2)a...(2)

subtracting eq1 from eq 2 we get

8=2a⇒a=4

usin this in eq 1 we get value of u=2

now for solving for the eighth second ,

S

8

=2+

2

1

×4×(2×8−1)=32m

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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

_______________________

Learn more :-

v= u+at

v²-u²= 2as

s = ut+1/2at²

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