Question

1. If a particle at rest starts moving in a horizontal straight line with uniform acceleration, what is the ratio of the distance covered during the 4th and the 3rd second.

2. What are the ratio of the distance travelled by a body, falling freely from rest in 1,2and 3 seconds.

Kindly give workings and then to give answer.

Answer 1

1).distance travelled during the nth second,
s= u + a/2[2t-1] ................(A)
for 4th second,
s4=0+a/2[2×4-1]
s4=7a/2...............(1)
SIMILARLY,
s3= 0+a/2[2×3-1]
s3=5a/2...................(2)
so,
s4/s3 = 7a/2 × 2/5a
s4:s3 = 7:5

2). when a body falls freely under gravity then the ratios of the distances covered is in the same ratio as the odd numbers are with each other.
for it , a=g, then again from (A),
s1 = 0+g/2[2×1-1]
s1=g/2........(1)
SIMILARLY, s2 =0+g/2[2×2-1]
s2=3×g/2
or,। s2= 3s1 from 1)
SIMILARLY, s3=0+g/2[2×3-1]
s3=5× g/2
or, s3=5s1 again from (1)
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sn=(2n-1)s1
hence,
s1:s2:s3:.................sn= 1:3:5:7:..............:(2n-1)

Answer 2

1)distance travelled during the nth second,

s= u + a/2[2t-1] ................(A)

for 4th second,

s4=0+a/2[2×4-1]

s4=7a/2...............(1)

SIMILARLY,

s3= 0+a/2[2×3-1]

s3=5a/2...................(2)

so,

s4/s3 = 7a/2 × 2/5a

s4:s3 = 7:5

2)s1 = 0+g/2[2×1-1]

s1=g/2........(1)

, s2 =0+g/2[2×2-1]

s2=3×g/2

or,। s2= 3s1 from 1)

, s3=0+g/2[2×3-1]

s3=5× g/2

or, s3=5s1 again from (1)

sn=(2n-1)s1

hence,

s1:s2:s3:.................sn= 1:3:5:7:..............:(2n-1)

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