The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
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Answer:
A velocity time graph shows the change of velocity of an object with respect ot time. If the slope of the graph is increasing in the postive region, it means that the velocity is changing, if the slope is decreasing, it means the the velocity is decreasing, but the object is moving in the same direction (positve direction).
If this slope intersects the graph at x-axis, it means that the body has 0 velocity and has become still. After that, if the line enters in the negative region, it means that its velocity is started to increases again, but the body is movinging in the opposite direction (negative direction)
