Question

a body leaving a certain point O with a constant acceleration at the end of 5th second its velocity is 1.5 metre per second. At end of 6th second the body stops and begins to move backwards. Find the distance transferred by the body before it stops determine the velocity with which the body returns to point O.​

Answer 1

Answer:

Given that:-

Let us consider

Intial Velocity = U m/s

Here,

Velocity gradually decreases. So acceleration a = -a m/s²

$\sf \red {At \:the \: end\: of \:5th\: second\: velocity}$

$\sf \pink \implies \purple {V = 1.5 \: m/s}$

$\sf \blue \implies \green {V = u + at}$

$\sf \red \implies \orange {1.5 = u + (-a)5}$

$\sf \purple \implies \pink {1.5 = u - 5a -(1) }$

At the end of 6th second, it's velocity is zero.

$\sf \green \implies \blue {i.e \: V = O \: m/s}$

$\sf \orange \implies \red {V = u + at}$

$\sf \pink \implies \purple {O = u + (-a)6}$

$\sf \blue \implies \green {O = u - 6a}$

$\sf \red \implies \orange {u = 6a -(2)}$

Substitute u = 6a in (1)

$\sf \purple \implies \pink {1.5 = 6a - 5a}$

$\sf \blue \implies \green {a = 1.5 \: m/s²}$

Substitute "a" Value in (2)

$\sf \orange \implies \red {u = 6(1.5)}$

$\sf \pink \implies \purple {u = 9 \: m/s}$

Now:-

Distance traversed by the body before it stops

$\sf \green \implies \blue {Total \: 6 \: seconds}$

$\sf \red \implies \orange {s = ut + \frac{1}{2} {at}^{2} }$

$\sf \purple \implies \pink {s = 9(6) + \frac{1}{2}( - 1.5)(6 {)}^{2} }$

$\sf \blue \implies \green {s = 54 - \frac{1}{2}(1.5)36}$

$\sf \orange \implies \red {s = 54 - 27}$

$\sf \pink \implies \purple {s = 27m}$

The body begins to move back from rest with a constant acceleration and reaches the point "O" with a velocity V.

$\sf \green \implies \blue {V² - u² = 2as}$

$\sf \red \implies \orange {V² - O² = 2(1.5)(27) }$

$\sf \purple \implies \pink {V² = 81}$

$\sf \blue \implies \green {V = \sqrt {81}}$

$\sf \orange \implies \red {V = 9 \: m/s \: in \: opposite\: direction}$