The graph of two cars that collide shows that they intersect at a point 5.5m on

Question

The graph of two cars that collide shows that they intersect at a point 5.5m on a distance/time graph. The first car has a velocity of 0.39m/s. The scond car has a velocity of 0.32m/s. The first line(first car) has a slope of 0.4m/s and the second line (second line) has a slope of -0.3m/s. The actual collision poin is 4.6m while the calculated collision point is 5.5m.

a) How do the two sets of values compare?

b) What does the steepness of the slope indicate?

c) Why is one slope positive and the other negative?

d) What are some reasons for the difference in the actual and calculated collision points?

Explanation / Answer

a) The first set : distance = 5.5 m

velocity of car 1 = 0.39 m/s

velocity of car 2 = 0.32 m/s

Second set :

The first set : distance = 4.6 m

velocity of car 1 = 0.4 m/s

velocity of car 2 = -0.3 m/s

The values are almost same...Except that in 1st set, the velocity of car 2, the direction must be negative...becuase the cars are assumed to be moving towards each other...HEnce one will have positive velocity and other will have negative velocity...

b)
The steepness of the slope gives the velocity...If the slope is high...means, the distance is changing very quickly with time...SO the velocity must be more...

The slope is given by : dx / dt = distance / time = velocity

c)

One slope is poitive and the other is negative because the cars are moving towards each other..so one will have a positive velocity and the other will have a negative velocity...

d) The reason for the different values is that, the calculations are based on certain assumptions, that the collision is either elastic or non elastic...Also, the loss of energy can not be calculated precisely...These account for the errors in the data...

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