To resolve point A and B we need the concepts related to conservation of momentum (By collision) and Kinetic Energy. Conservation of momentum is given by the equation,

Our values in the statment are:




Part A) As it is in an icy intersection, there is two different components (x,y) then,



Then the magnitude is,

Part B) To obtain the Kinetic Energy Loss we need to use its equation, which is given by,



The final energy is given by,



Then the change in Kinetic Energy is


<em>There was a loss of KE of 183.02kJ</em>
Answer:
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Answer:
For a velocity versus time graph how do you know what the velocity is at a certain time?
Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.
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How do you know the acceleration at a certain time?

Hence,
By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.
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How do you know the Displacement at a certain time?
Ans: As Displacement equals to the area enclosed by the slope of the Velocity-Time Graph, By finding the area under the slope till the perpendicular at the desired time, we find the Displacement.
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For this question we should apply
a = v^2 - u^2 by t
a = 69 - 0 by 4.5
a = 69 by 4.5
a = 15.33
a = 6.85 m/s^2
If the answer in option is near to answer then , you can mark it as correct.
.:. The acceleration is 6.9 m/s^2