Question

A body leaving a certain point 'O' moves with an a constant acceleration. At the end of the 5 th second its velocity is 1.5 m/s. At the end of the sixth second the body stops and then begins to move backwards. Find the distance traversed by the body before it stops. Determine the velocity with which the body returns to point 'O' ? (27m,-9m/s)

Answer 1

at the end of the 5th sec its velocity is 1.5 m/s
let intial velocity is u and acceleration is a
e.g., 1.5 = u + 5a ----(1)
at the end of 6th body stops.
e.g., 0 = u + 6a -----(2)

from equations (1) and (2),
1.5 = -6a + 5a => a = -1.5 m/s²
now, put a in equation (1),
1.5 = u + 5 × (-1.5)
u = 1.5 + 7.5 = 9 m/s

distance travelled by the body before it stop
v² = u² + 2aS
0 = 9² + 2(-1.5)S
S = 81/3 =27 m
now, velocity at 'O' is v
use formula, v² = u² + 2aS
v² = 0² + 2(1.5) × 27
v² = 2 × 1.5 × 27 = 81
v = 9m/s
hence, velocity of body to return at 'O' is -9m/s

Answer 2

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.

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

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

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

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

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