Physics, asked by captaingen1735, 3 months ago

A particle moves with uniform acceleration along a straight line. it covers 41 cm and 49 cm in the 6th and 10th sec. what will be the distance covered by the particle in 15th sec.

Answers

Answered by Anonymous
10

Answer:

The distance travelled during the nth second of motion of a body is given by

Sñ = u + 1/2 a(2n-1)

for the motion of during the 8th second,

3 = u + 1/2 a(16 - 1) = u + 15a/2 ..........(I)

for the motion during the 16th second,

5 = u + 1/2 a(32 - 1) = u + 31a/2 ...........(ii)

Now,

subtracting equation (I) from (ii)

8a = 2

acceleration (a) = 1/4 m/sec²

from equation (I) , u = 3 - (15/2 * 1/4 ) = 9/8 m/sec

Now,

the velocity at the end of 5 second

V1 = u + 5a

average velocity during this interval of 10 seconds

= v1 + v2/2

= (u + 5a) + (u + 15a)/2

= u + 10 a

distance travel during the interval

S = average velocity * time

= ( u + 10a ) * t

= (9/8 + 10/4) * 10

= 290/8

= 36.25m...

hope it helps

Answered by SuvadipSingha
2

Explanation:

The distance travelled in n

th

second is

S

n

=u+

2

1

(2n−1)a ....(1)

So distance travelled in t

th

&(t+1)

th

second are

S

t

=u+

2

1

(2t−1)a ....(2)

S

t+1

=u+

2

1

(2t+1)a ....(3)

As per question,

S

t

+S

t+1

=100=2(u+at) ....(4)

Now from first equation of motion the velocity of particle after time t, if it moves with an acceleration a is

v=u+at ....(5)

where u is initial velocity

So from equation (4) and (5), we get v=50cm/s

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