Question

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

Answer 1

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.

Answer 2

Answer

• Distance travelled in 8th second = 0 m

Given

• a body starts with initial velocity 30 m/s and retardation 4 m/s²

To Find

• Distance travelled in 8th second

Concept Used

• Equation of motion

$\bigstar \bf \;\; S_n=u+\dfrac{a}{2}[2n-1]$

Solution

Initial velocity , u = 30 m/s

Retardation , a = - 4 m/s²

n th second , n = 8

Apply formula .

$\implies \rm S_n=u+\dfrac{a}{2}[2n-1]\\\\\implies \rm S_8=30+\dfrac{-4}{2}[2(8)-1]\\\\\implies \rm S_8=30-2[15]\\\\\implies \rm S_8=30-30\\\\\implies \rm S_8=0\ m$

More Info

$\boxed{\begin{minipage}{4.5cm} \underbrace{\bf{Equations\ of\ motion}}\ :\\\\\rm \bigstar \;\; v=u+at \\\\\rm \bigstar \;\; s=ut+\dfrac{1}{2}at^2\\\\\rm \bigstar \;\; v^2-u^2=2as\\\\\rm \bigstar \;\; S_n=u+\dfrac{a}{2}[2n-1] \end{minipage}}$