SOLUTION: Hey again! This is also impossible, i can do question 3 but not 2! :D A racing car is moving with constant acceleration along a straight stretch of road. The car passes point A

Question 107896This question is from textbook Mechanics I
: Hey again! This is also impossible, i can do question 3 but not 2! :D
A racing car is moving with constant acceleration along a straight stretch of road. The car passes point A at 0s, travelling at 18ms - 1, point B at 3s and point C at 7s, travelling at 53ms - 1. Calculate:
A - acceleration of the car
B - Speed of the car at B
C - the distance from B to C.
I've tried all three parts and failed miserably as i have the answers in the back of the book, i have no clue! :( Thanks alot in advance! :D This question is from textbook Mechanics I

Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
A) constant acceleration is (change in velocity)/(change in time) ... (53-18)/(7-0)

B) velocity=(initial velocity)+(acceleration*time) ... Vb=18+3a

C) distance=(average velocity)*(time) ... d=((53+Vb)/2)*(7-3)
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