Question

QUESTION IN ATTACHMENT

Answer 1

APPROPRIATE QUESTION:-

[THE CORRECT QUESTION IS TILL THIRD LINE]

EXPLANATION:-

Let u be the initial velocity ,a be the acceleration of the particle.Distance covered by the particle in 8th second is 3 m .Using the equation of s_(nth)

Putting values:-

3=ut+1/2a(2×8-1)

3=u+1/2(15a)

→2u+15a=6 ---eq-1

Similarily

5=u+1/2(2×16-1)

5=u+1/2(31a)

2u+31a=10--eq-2

EQ-2-EQ-1

2u+31a-(2u+15a)=10-6

16a=4

a=4/16=1/4 m/s²

Using equation-q to get u:-

2u+15a=6

2u+15×1/4=6

u=9/8 m/s

So we can easily find the final velocity covered:-

v=u+at

v=9/8+1/4×5

v=19/8 m/s

Now let's calculate the distance:-

S=ut+1/2at²

S=19/8(10)+1/2(1/4)(10)²

S=190/8+100/8

S=290/8

S=36.25 m