Question

a body starts from rest with uniform acceleration of 2m/s2 until it travels a distance of 625 meter find its velocity​

Answer 1

Answer:

Given:

Initial velocity = 0 m/s

Acceleration = 2 m/s²

Distance travelled = 625 m

To find:

Final Velocity

Calculation:

${v}^{2} = {u}^{2} + 2as$

$= > {v}^{2} = {0}^{2} + (2 \times 2 \times 625)$

$= > v = \sqrt{(2 \times 2 \times 625)}$

$= > v = 2 \times 25$

$= > v = 50 \: m \: {s}^{ - 1}$

So final answer :

$\boxed{ \red{final \: vel. = 50 \: m \: {s}^{ - 1}}}$

Additional information about velocity:

1. It is a vector quantity whereas speed is a scalar quantity.

2. It consists of both magnitude and direction

3. Average Velocity is the ratio of total displacement to the total time taken.

Answer 2

Given :

• Initial Velocity = 0 m/s
• Acceleration = 2 m/s²
• Distance covered = 625 m

To Find :

• Velocity (v)

Solution :

Solving via first calculating the time using second equation of motion and then using third equation of motion to find final velocity, just a little different approach ^^"

We have,

• s = distance = 625 m
• u = 0 m/s
• t (to be calculated)
• a = 2 m/s²

So this gives the perfect set for using second equation of motion.

Third Equation of Motion :

$\bold{\large{\boxed{\mathtt{s=\:ut\:+\:{\dfrac{1}{2}\:\:at^2}}}}}$

Block in the available data,

➠ $\mathtt{625=0\:\times\:t\:+\:{\dfrac{1}{2}\:2\:t^2}}$

➠ $\mathtt{625=0\:+\:{\dfrac{1}{2}\:2t^2}}$

➠ $\mathtt{625\:=\:{\dfrac{1}{2}\:\:2t^2}}$

➠ $\mathtt{625=\:{\dfrac{2t^2}{2}}}$

➠ $\mathtt{1250\:=\:2t^2}$

➠ $\mathtt{\dfrac{1250}{2}\:=\:t^2}$

➠ $\mathtt{625\:=\:t^2}$

➠ $\mathtt{\sqrt{625}\:=\:t}$

➠ $\mathtt{25\:=t}$

$\bold{\sf{\therefore{\underline{time\:taken\:=\:25\:second}}}}$

Now, simply use the third equation of motion and calculate v.

$\bold{\large{\boxed{\mathtt{v\:=\:u\:+\:at}}}}$

Block in the available data,

➠ $\mathtt{v=0\:+\:2\:\times\:25}$

➠ $\mathtt{v=\:50}$

$\bold{\sf{\therefore{\underline{Final\:Velocity\:(v)\:=\:50\:ms^-1}}}}$