Question

An object is moving along the x - axis with uniform acceleration of 4m/s^2. At time t=0,X = 5 m
and v= 3 m/s.
(a) What will be the velocity and the position of the object at time t = 25.
(b) What will be the position of the object when it has velocity of 5 m/s.​

Answer 1

$\mathfrak{\large{\underline{\underline{Answer:}}}}$

a)The Velocity and the positions of the object at t=2 is 11m/s and 19m.

b)The position of the object is 7m.

Given:

Acceleration a = $\4m/{s}^{2}$

Velocity v = 3m/s

Time t = 0 s

Position x = 5m

$\mathfrak{\large{\underline{\underline{Explanation}}}}$

(a) The Velocity and the positions of the object at t = 2 s.

The Velocity after 2 s will be,

Using the equation of motion:

$v = u + at$

Put the values in the equation.

$v = 3 + 4 \times 2 \\ v = 11m \: s$

Now, The distance travelled after 2 sec.

Using third equation of motion:

$v {}^{2} - u {}^{2} = 2as \\ \frac{v {}^{2} - u {}^{2} }{2as} = s \\ s = \frac{11 {}^{2} - 3 {}^{2} }{2 \times 1} \\ s = 14m$

So, the position will be

$x \frac{}{0} = x + s \\ x \frac{}{0} = 5 + 14 \\ x \frac{}{0} = 19m$

(b) The position of the object when it has Velocity of 5 m/s.

$v {}^{2} - u {}^{2} = 2as \\ s = \frac{v {}^{2} - u {}^{2} }{2a} \\ s = \frac{5 {}^{2} - 3 {}^{2} }{2 \times 4} \\ s = 2m$

So, the position will be:

x' = x + s

x' = 5 + 2

x' = 7 m.

$\mathfrak{\large{\underline{\underline{Verification}}}}$

(a) The velocity and the position of the object at t = 2 s is 11 m/s and 19 m.

(b) The position of the object is 7 m.

Answer 2

Answer :

a)The Velocity and the positions of the object at t=2 is 11m/s and 19m.  

b)The position of the object is 7m.

Explanation :

Given that :

Acceleration (a) = 4m/s²

Velocity (v) = 3m/s

Time taken (t) = 0 s

Position (y) = 5m.

To Find :

(a) What will be the velocity and the position of the object at time t = 25.

Solution :

As given,

Velocity and the positions of the object at t = 2 s.

The velocity be :

We know that :

Equation of Motion :

$\bigstar\;\boxed{\sf v = u+at}}$

Plug the given values :

$v = 3 + 4 \times 2\\\\ v = 11m\s.$

Let the position be y,

Hence,

The velocity is 11m/s.

We also know that :

$\bigstar\;\boxed{\sf v^2-u^2= 2as}}$

Put the given values :

$\sf \implies \dfrac{v^2-u^2}{2as} = s\\\\ \sf \implies s = \dfrac{11^2-3^2}{2\times 1}\\\\ \sf \implies s = 14m$

Position will be :

$\sf \implies y = x + s\\\\\sf \implies y = 5 + 14\\\\\ \sf \implies y = 19m.$

Hence,

The  positon is 19m.

Now,

To Find :

What will be the position of the object when it has velocity of 5 m/s.​

We also know that :

$\bigstar\;\boxed{\sf v^2-u^2= 2as}}$

$\sf \implies \dfrac{v^2-u^2}{2as} = s\\\\\sf \implies s = \dfrac{5^2-3^2}{2\times 4}\\\\ \sf \implies s = 2m$

Let the positon be x'

So,

The positon will be:

$\sf \implies x' = x + s\\\\\implies x' = 5 + 2\\\\\implies x' = 7 m.$