Question

a body starts from rest with a uniform acceleration 2 ms^ -2. Find the distance covered by the body in 2 s​

Answer 1

Answer:

4m

Explanation:

a= 2 ms^-2

t=  2s

u=0

s=?

s=ut+1/2at^2

=1/2*2*2

=4m

Answer 2

GIVEN:-

• $\rm{Initial\:Velocity=0m/s}$

• $\rm{Acceleration=2m/s^-2}$

• $\rm{Time\:Taken=2s.}$.

TO FIND:-

• The Distance Covered by the Body.

FORMULAE USED:-

• ${\boxed{\rm{S=ut+\dfrac{1}{2}\times{a}\times{(t)^2}}}}$

Where,

U= initial velocity

a = acceleration

S= Distance

t= time

Now,

$\implies\rm{S=ut+\dfrac{1}{2}\times{a}\times{(t)^2}}$

$\implies\rm{S=0\times{t}+\dfrac{1}{\cancel{2}}\times{\cancel{2}}\times{4}}$

$\implies\rm{S=0+4}$

$\implies\rm{S=4m}$.

Hence, The Body travelled 4m.

Some more equation of motion

• ${\boxed{\rm{\red{V=u+at}}}}$

• ${\boxed{\rm{\blue{v^2-u^2=2as}}}}$.