Question

A body is moving with a constant acceleration of 4m/s×s covers a distance of 15m during 3rd second of its motion .calculate the velocity of the body?

Answer 1

The body covers 15 m displacement in the 3rd second. Acceleration is given as 4 m/s².

We have to find the :

• Initial velocity of body

• Velocity at 3 sec

Calculation:

We shall use the relationship between displacement at n th second ;

Let initial velocity be u

$d = u + \frac{1}{2} a(2n - 1)$

$= > 15 = u + \frac{1}{2} \times 4 \times \{(2 \times 3 )- 1 \}$

$= > 15 = u + 10$

$= > u = 5 \: m {s}^{ - 1}$

Now velocity at 3 secs be v ;

$\therefore \: \: v = u + at$

$= > v = 5 + (4 \times 3)$

$= > v = 17 \: m {s}^{ - 1}$

Answer 2

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QUESTION -

A body is moving with a constant acceleration of 4m/s×s covers a distance of 15m during 3rd second of its motion .calculate the velocity of the body?

SOLUTION -

We know that for the distance travelled by a body in the n the second is :

$\sf{\orange{\leadsto{\boxed{\boxed{ S_{n} = u + \dfrac{1}{2} \times a \times ( 2n - 1 ) }}}}}$

Where :

S_{n} is the distance travelled by the body in the nth second.

U is the innitial Velocity of the body.

A is the acceleration of the body.

Now, let us substitute the given Values in the Equation :

$\sf{\red{\leadsto{\boxed{\boxed{ S_{3} = u + \dfrac{1}{2} \times 4 \times5 = 15 m }}}}}$

$=> u + 10 = 15 => u = 5 m / s$

So, the innitial Velocity of the body is 5 m / s.

Now using First Equation of motion :

=> V = u + at

Where :

V is the final Velocity of the body.

U is the innitial velocity of the body.

A is the acceleration of the body.

T is the time taken.

Substuting the required values, the Equation becomes :

$V = 5 + 4 \times 3 = 5 +12 = 17 \: m/s.$

So the Velocity of the body during the third second of it's motion becomes 17 m / s.

ANSWER -

the Velocity of the body during the third second of it's motion becomes 17 m / s.