Asked by Grace
A,B,C are three points on a straight road.A car passes A with speed of 5m/s and travel from A to B with instant acceleration of 2m/s^2.from B to C has a constant retardation of 3.5m/s^2 and comes to rest at c.i.e the total distance from A to C is 475.find the speed of the car at B,and the distance of B from A
Answers
Answered by
oobleck
let's say it is x meters from A to B, and thus 475-x meters from B to C
at A, v = 5
If it takes t1 seconds to travel to B, then v = 5+2t1 at B
and x = 5t1 + t1^2
to get to C, if it takes t2 seconds, then we have
(5+2t1)t2 - 1.75 t2^2 = 475-(5t1+t1^2)
5+2t1 - 3.5t2 = 0
solve those last two equations, and you get
t1 = 15 and t2 = 10
so AB = x = 5*15 + 15^2 = 300 meters
at A, v = 5
If it takes t1 seconds to travel to B, then v = 5+2t1 at B
and x = 5t1 + t1^2
to get to C, if it takes t2 seconds, then we have
(5+2t1)t2 - 1.75 t2^2 = 475-(5t1+t1^2)
5+2t1 - 3.5t2 = 0
solve those last two equations, and you get
t1 = 15 and t2 = 10
so AB = x = 5*15 + 15^2 = 300 meters
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