Question

A body travels 20m in the 2nd second and 40m in 4th second.Find actual length covered by the body in 10s

Answer 1

Answer:

Distance travelled = 50 m

Explanation:

given that,

A body travels 20m in the 2nd second

we know that,

when the body travels a distance in nth second

then distance travelled S1

S1 = u + a/2(2n - 1)

where,

u = initial velocity

a = acceleration

so

S2 = u + a/2(2(1) - 1) = 20

= u + a/2(2 - 1) = 20

= u + a/2 = 20

2u + a = 40. ...(1)

now,

It also traveled 40 m in 4th second.

so,

S4 = u + a/2(2(4) - 1) = 40

= u + a/2(8 - 1) = 40

u + a/2 × 7 = 40

u + 7a/2 = 40

2u + 7a = 80. ...(2)

now,

we have,

2u + a = 40. ...(1)

2u + 7a = 80. ...(2)

substracting (2) from (1)

2u + a - (2u + 7a) = 40 - 80

2u + a - 2u - 7a = -40

-6a = -40

a = -40/-6

a = 20/3 m/s²

now,

putting the value of a on (1)

2u + a = 40

2u + 20/3 = 40

2u = 40 + 20/3

2u = 100/3

u = 100/3 × ½

u = 50/3 m/s

now,

we have,

initial velocity(u) = 50/3 m/s

acceleration(a) = 20/3 m/s

given time(t) = 10 s

so,

from the equation of motion,

S = ut + ½at²

putting the values,

S = 50/3 × 10 + ½ × 20/3 × 10 × 10

S = 500/3 + 1000/3

S = 1500/3 m

S = 50 m

so,

Distance travelled = 50 m

Answer 2

Answer:

Answer:Distance travelled = 50 m

mExplanation:

REFER TO THE ATTACHMENT

SORRY FOR THE BAD HANDWRITING