Question

A body started from rest and moved a distance of 1km at a uniform asscilleration of 5m/s . find out the time taken for this motion.​

Answer 1

Answer

Time taken = 20 s

Given

A body started from rest and moved a distance of 1 km at a uniform acceleration of 5 m/s

To Find

Time taken throughout the motion

Concept Used

As the acceleration of the body is uniform throughout the motion , so we need to apply equation's of kinematics .

2nd equation of motion

→ s = ut + ¹/₂at²

Solution

Initial velocity , u = 0 m/s

[ ∵ starting from rest ]

Distance , s = 1 km = 1000 m

Acceleration , a = 5 m/s²

Time , t = ? s

Apply 2nd equation of motion ,

⇒ s = ut + ¹/₂at²

⇒ (1000) = (0)t + ¹/₂ (5)t²

⇒ 1000 = 0 + ¹/₂(5)t²

⇒ 2000 = 5t²

⇒ 5t² = 2000

⇒ t² = 400

⇒ t = 20 s

So , time taken for it's motion is 20 s

Answer 2

QUESTION :-

A body started from rest and moved a distance of 1km at a uniform acceleration of 5m/s² . Find out the time taken for this motion.

GIVEN :-

• A body is started from rest and covered a distance of 1km with uniform accleration of 5m/s²

TO FIND :-

• Time taken for this motion

SOLUTION :-

u = 0 (Since the body has started from rest)

a = 5m/s²

s = 1 km = 1000m

From second equation of motion ,

$\large\rm\bold{\boxed{S\:=\:ut+\frac{1}{2}\:at^2}}$

Where,

• S is distance covered by the body
• u is intial velocity of body
• t is time taken by the body
• a is acceleration of body

$\large\rm{\rightarrow{1000\:=\:0(t)+\frac{1}{2}\:(5)(t^2)}}$

$\large\rm{\rightarrow{1000\:=\:\frac{5t^2}{2}}}$

$\large\rm{\rightarrow{2\times\:1000\:=\:5t^2}}$

$\large\rm{\rightarrow{2000\:=\:5t^2}}$

$\large\rm{\rightarrow{t^2\:=\:\frac{2000}{5}}}$

$\large\rm{\rightarrow{t^2\:=\:400}}$

$\large\rm{\rightarrow{t\:=\:\sqrt{400}\:=\:20\:sec}}$

∴ The time taken by the body for this motion is 20 sec

$\huge\rm{\underline{\green{Additional\:Info:-}}}$

❃ First equation of motion is given by ,

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

Where ,

• v is final velocity of particle
• u is initial velocity of particle
• t is time taken
• a is acceleration of particle

❃ Third equation of motion is given by,

$\large\rm\bold{\boxed{v^2-u^2\:=\:2as}}$

Where ,

• v is final velocity of particle
• u is initial velocity of particle
• a us acceleration of particle
• s is distance covered

$\huge\rm\underline{Note:-}$

Equations of Motion are applicable only when the acceleration is constant