A car moving with constant acceleration covers the distance between two points 180m apart in 6sec.its speed as it passes the second point is 45m/s.what is its acceleration and its speed at the first point.

Answer 1

measuring from the first point,
s(t) = vt + 1/2 at^2
we know that s(6) = 180, so
6v + 18a = 180
Also, we have
v + 6a = 45

So, v=15 and a = 5

Answer 2

Acc. to the question v=u+at ;v=45m/s , t=6s ;45=u+6a ;(45-u)/6=a----1 ; v^2-u^2=2as ; 2025-u^2=360a ---2 ; subsitute --1 in ---2 eq ; 2025-u^2=360*(45-u)/6 by solving we will get a equation ; u^2-u+675 ;we will get u=15 or 45 ; we have to take 15 =u; then v=u+at ; 45=15+6a ; a=5m/s

Answer 3

Acc. to the question v=u+at ;v=45m/s , t=6s ;45=u+6a ;(45-u)/6=a----1 ; v^2-u^2=2as ; 2025-u^2=360a ---2 ; subsitute --1 in ---2 eq ; 2025-u^2=360*(45-u)/6 by solving we will get a equation ; u^2-u+675=0 ;we will get u=15 or 45 ; we have to take 15 =u; then v=u+at ; 45=15+6a ; a=5m/s

Answer 4

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Answer 5

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