Question

A car initially has a velocity of 29 m/s starts to increase its velocity at a constant rate of 2.4 m/s2, for a distance of 22 m. What is its final velocity?

Answer 1

vf=43.55m/s

Step-by-step explanation:

Data:

vi=29m/s

S=22m

a=2.4 m/s^2

vf=? we are supposed to find final velocity.

we will use following equation of motion:

Formula:

`2as=vf^(2) -vi^(2)`

put values

`2(2.4)(22)=vf^2-(29)^2\\`

solving for
`vf^2`.

taking
`(29)^2` on the other side .

`vf^2=2(2.4)(22)+(29)^2`

solve square first.

`vf^2=2(24)(22)+841`

multiply the terms.

`vf^2=1056+841`

add the terms

`vf^2=1897`

taking square root on both sides.

`√(vf^2)=√(1897)`

square root of
`vf^2` will be cancelled out .

`vf=43.55m/s`