Question

Velocity of a body change from 10m/s to ŕ0m/s in 2s find the acceleration and the distance in 2s pzz

Answer 1

Given u (initial velocity) = 10m/s
v (final velocity) = 40m/s
t = 2s

a =
$\frac{v - u}{t}$
a =
$\frac{40 - 10}{2} \\ = \frac{30}{2} \\ = 15$
acceleration is 15m/s²


Distance (s) = ut + 1/2at²
=
$10 \times 2 + \frac{1}{2} \times 15 \times {2}^{2} \\ = 20 + \frac{1}{2} \times 15 \times 4 \\ = 20 + 30 \\ = 50$
So, acceleration is 15m/s² and distance is 50m.


Hope it helps dear friend ☺️

Answer 2

◢GIVEN

● INITIAL VELOCITY = u = 10 m/s

● FINAL VELOCITY = v = 40 m/s

◢AND

● TIME = t = 2 sec

______________________________

● WE KNOW THAT :-

=> v = u + at

$= > a = \frac{v - u}{t} \: \: \: ....Eq _{1}$
______________[a=Acceleration]

● PLUG THE VALUE OF "v", "u" and "t"

◢WE GET

$= > a = \frac{40 - 10}{2} = \frac{30}{2} \\ \\ = > a = 15 \: \frac{m}{ {s}^{2} }$
____________________[◢ANSWER]

◢NOW

● WE HAVE ACCELERATION = a = 15 m/s²

________________________________

● DISTANCE = s

◢AND

$= > s = ut \: + \: \frac{1}{2} a {t}^{2} \\ \\ = > s = 10 \times 2 + \frac{1}{2} \times 15 \times {2}^{2} \\ \\ = > s = 20 + 30 \\ \\ = > s \: = 50 \: m$
____________________[◢ANSWER]

◢HENCE,

● ACCELERATION = 15 m/s²

◢AND

● DISTANCE = 50 m

=====================================

☆$BE \: \: BRAINLY$☆