Question

A body starts from rest and moves with a unifornm acceleration of 5m/s square a body starts from rest and moves with a uniform acceleration of 5 M per second square what will be its velocity displacement at the end of 12 second

Answer 1

v= u+ at
v= 0+ 5×12
v= 0+ 60
v= 60m/s

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Final\:velocity=60\:m/s}}}$

$\green{\therefore{\text{Displacement=360\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a body starts from rest and moves with a unifornm acceleration of 5m/s^2.

• We have to find the final velocity and displacement of body.

$\green{\underline\bold{Given :}} \\ :\implies \text{Initial\:velocity(u)=0\:m/s} \\ \\ :\implies \text{Acceleration(a)= 5\:m/s}^{2}\\ \\ :\implies \text{Time(t)= 12\:sec}\\ \\ \red{\underline \bold{To \: Find : }} \\ :\implies \text{Final\:velocity(a)= ?}\\\\ :\implies Distance(s)=?$

• According to given question :

$\bold{Using \: first\:equation\:of\:motion } \\ : \implies v = u + at \\ \\ : \implies v= 0+5\times 12 \\ \\ \green{ : \implies{ \text{v=60\:m/s}}}$

$\bold{Using \: second \: equation \: of \: motion } \\ : \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ : \implies s = 0 \times 12 + \frac{1}{2} \times 5 \times {12}^{2} \\ \\ : \implies s = \frac{1}{2} \times 5 \times 144 \\ \\ : \implies s =72 \times 5 \\ \\ \green{ : \implies \text{ s =360 \: m}}$