Question

A body starting from rest attains a velocity of 40cm/second in 8 second. Determine the velocity of the body at the end of 14 second...

Answer 1

u=0
V=40cm/s=0.4m/s
t=8s
a=5m/s
t=14s
v=u+at
=110m/s

Answer 2

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

$\green{\therefore{\text{Velocity}_{(\text{v\:at\:14\:sec}})=70\:cm/s}}$

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

• In the given question information given about a body starting from rest attains a velocity of 40cm/second in 8 second.

• We have to find velocity of the body at the end of 14 second.

$\green{\underline\bold{Given :}} \\ :\implies \text{Initial\:velocity(u)=0 m/s} \\ \\ :\implies {\text{velocity}_{\text{(v\:at\:t=8\:sec}})= 40 \:cm/s}\\ \\ :\implies \text{Time= 8\:sec}\\ \\ \red{\underline \bold{To \: Find : }} \\ :\implies {\text{velocity}_{\text{(v\:at\:t=14\:sec)}}= ?}$

• According to given question :

$\bold{Using \: first \: equation \: of \: motion : } \\ : \implies v = u + at \\ \\ : \implies 40= 0 + a \times 8 \\ \\ : \implies 8\times a = {40} \\ \\ : \implies a= \frac{40}{8} \\ \\ \green{ : \implies {\text{a = 5 \: cm/s}^{2}}}$

$\text{Velocity\:at\:14\:sec}\\\\ \bold{Using \: first \: equation \: of \: motion : } \\ : \implies v = u + at \\ \\ : \implies v= 0 + 5\times 14 \\ \\ \green{ : \implies {\text{v=70 \: cm/s}}}$