Question

a body travels from rest with a uniform acceleration of 7ms^2. Determine the velocity of the body when it covers a distance of 25m?

Answer 1

Given-

• Initial velocity= 0m/s
• Acceleration= 7m/s²
• Distance covered = 25m

   

Find-

• Final Velocity-?

     

Formula to be used-

$\boxed{\tt {v^2-u^2= 2as}}$

where,

v= Final Velocity

u= Initial velocity

a= Acceleration

s= Distance Covered

   

Solution-

By using 3rd equation of motion,

$\bf{v^2-u^2= 2as}$

$\bf{ \implies v^2 -0= 2 \times 7 \times 25 }$

$\bf{\implies v^2 = 14 \times 75}$

$\bf{\implies v^2 = 1050}$

$\bf{ \implies v= \sqrt{1050} }$

$\bf{\implies v=5 \sqrt{42} }$

$\bf{\implies v= 32.40 m/s}$

   

∴ The velocity of the body when it covers a distance of 25m is 32.40m/s.

Answer 2

Answer:

Given-

Initial velocity= 0m/s

Acceleration= 7m/s²

Distance covered = 25m

   

Find-

Final Velocity-?

     

Formula to be used-

\boxed{\tt {v^2-u^2= 2as}}v2−u2=2as

where,

v= Final Velocity

u= Initial velocity

a= Acceleration

s= Distance Covered

   

Solution-

By using 3rd equation of motion,

\bf{v^2-u^2= 2as}v2−u2=2as

\bf{ \implies v^2 -0= 2 \times 7 \times 25 }⟹v2−0=2×7×25

\bf{\implies v^2 = 14 \times 75}⟹v2=14×75

\bf{\implies v^2 = 1050}⟹v2=1050

\bf{ \implies v= \sqrt{1050} }⟹v=1050

\bf{\implies v=5 \sqrt{42} }⟹v=542

\bf{\implies v= 32.40 m/s}⟹v=32.40m/s

   

∴ The velocity of the body when it covers a distance of 25m is 32.40m/s.