1.

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»


Solution :`V^(2)-u^(2) =2axx180 rArr 45^(2)-u^(2) =360 a`
`v=u+at rArr 45=u+axx6`
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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`.


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