Explanation:
Given that,
- Distance travelled by the car in first second = 8 m
⠀⠀⠀⇒ S′ = u + a/2(2n ― 1)
⠀⠀⠀⇒8 = u + a/2× [2(1) ― 1]
⠀⠀⠀⇒8 = u + a/2× [2 ― 1]
⠀⠀⠀⇒8 = u + a/2× 1
⠀⠀⠀⇒8 = u + a/2⠀⠀⠀⠀⠀ … ( 1 )
And,
- Distance travelled by the car in second second = 12 m
⠀⠀⠀⇒ S″ = u + a/2(2n ― 1)
⠀⠀⠀⇒12 = u + a/2× [2(2) ― 1]
⠀⠀⠀⇒12 = u + a/2× [4 ― 1]
⠀⠀⠀⇒12 = u + a/2× 3
⠀⠀⠀⇒12 = u + 3a/2⠀⠀⠀⠀ … ( 2 )
Subtract ( 1 ) from ( 2 ).
⠀⠀⠀⇒12 ― 8 = u + 3a/2 ― (u + a/2)
⠀⠀⠀⇒4 = u + 3a/2 ― u ― a/2
⠀⠀⠀⇒4 = 3a/2 ― a/2
⠀⠀⠀⇒4 = 2a/2
⠀⠀⠀⇒4 × 2 = 2a
⠀⠀⠀⇒8 = 2a
⠀⠀⠀⇒8 ÷ 2 = a
⠀⠀⠀⇒4 m/s² = a⠀⠀⠀⠀⠀[Ans]
Substitute the value of a in equation ( 1 ).
⠀⠀⠀⇒8 = u + a/2⠀⠀⠀⠀⠀ … ( 1 )
⠀⠀⠀⇒8 = u + 4/2
⠀⠀⠀⇒8 = u + 2
⠀⠀⠀⇒8 ― 2 = u
⠀⠀⠀⇒6 m/s = u⠀⠀⠀⠀⠀[Ans]
Now, distance travelled in third second.
⠀⠀⠀⇒ S′″ = u + a/2(2n ― 1)
⠀⠀⠀⇒ S′″ = 6 + 4/2 × [2(3) ― 1]
⠀⠀⠀⇒ S′″ = 6 + 2 × [6 ― 1]
⠀⠀⠀⇒ S′″ = 6 + 2(5)
⠀⠀⠀⇒ S′″ = 6 + 10
⠀⠀⠀⇒ S′″ = 16 m⠀⠀⠀⠀⠀[Ans]
