A car starts from rest and moves with constant acceleration and covera the distance between two point `180 m` apart in `6 s`. Itsspeed as it passes the second point is `45 ms^(-1)` Find
a. Its acceleration
b. Its speed when it was at the point
c. The distance from the first point when it was at rest`.

Answer

a. `5 m s^(-2);
b. `15 ms^(-1);
c. 22.5 m`

Explanation

`v^(2)-u^(2) =2axx180 rArr 45^(2)-u^(2) =360 a`
`v=u+at rArr 45=u+axx6`
.
Solve to get `u=15 m s^(-1)`, `a= 5 m s^(-2)`
`u^(2)=0^(2)+2ax rArr 15^(2)=0^(2) + 2 xx5x rArr `x=22.5 m`.