Question

a body starts with initial velocity 30m/s and retardation 4 m/s^2.find distance travelled in 8th s​

Answer 1

Joshi sir comment

by v = u - at

0 = 30 - 4t implies t = 7.5

it means velocity will become 0 in 7.5 sec.

distance covered in 7.5 sec. = 30*7.5 - 1/2*4*(7.5)2 = 225 - 225/2 = 225/2

distance covered in 7 sec. = 30*7 - 1/2*4*(7)2= 210 - 98 = 112

so distance covered in first half of eighth sec. = 112.5 - 112 = 0.5 m

distance covered in last half of eighth sec. = 0*0.5 + 1/2*4(0.5)2= 0.5

so total distance covered in eighth sec. = 0.5 + 0.5 = 1 meter

Answer 2

☞Answer:

The distance travelled in 8ᵗʰ sec = 0 m

☞GIVEN:

Initial velocity = 30m/s.

Retardation = 4 m/s²

☞TO FIND:

The distance travelled in 8ᵗʰ second.

☞EXPLANATION:

Sₙ = u + a/2(2n - 1)

u = 30 m/s

a = - 4 m/s²

n = 8

S₈ = 30 - 4/2(2(8) - 1)

S₈ = 30 - 2(16 - 1)

S₈ = 30 - 2(15)

S₈ = 30 - 30

S₈ = 0 m

HENCE THE DISTANCE TRAVELLED AT EIGHTH SECOND WILL BE ZERO.