Asked by ronit
                a car starting from rest acquired a speed of 25ms-¹ in 10sec,after which it maintain this speed for 10sec. find (a)the acceleration (b) distance traveled during acceleration (c)total distance traveled?
            
            
        Answers
                    Answered by
            Steve
            
    acceleration is (25m/s)/(10s) = 2.5 m/s^2
distance during acceleration: s = 1/2 at^2 = 125m
so, total distance is 125 + 10*25 = 375 m
    
distance during acceleration: s = 1/2 at^2 = 125m
so, total distance is 125 + 10*25 = 375 m
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.