SOLUTION: A car starts from rest and accelerates at 1ms^{-2} for 10 seconds It then continues at uniform speed for another 20 seconds and decelerates to rest in 5 seconds.Find the: (a)dist

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Question 1207778: A car starts from rest and accelerates at 1ms^{-2} for 10 seconds It then continues at
uniform speed for another 20 seconds and decelerates to rest in 5 seconds.Find the:
(a)distance travelled;
(b) average speed;
(c) time taken to cover half the distance.
Answer by ikleyn(52925) About Me  (Show Source):
You can put this solution on YOUR website!
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A car starts from rest and accelerates at 1ms^{-2} for 10 seconds It then continues at
uniform speed for another 20 seconds and decelerates to rest in 5 seconds. Find the:
(a)distance travelled;
(b) average speed;
(c) time taken to cover half the distance.
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(a)  To find the length of the accelerated part (or path), use the formula

        d%5B1%5D = %28a%5Bacc%5Dt%5E2%29%2F2 = %281%2A10%5E2%29%2F2 = 100%2F2 = 50 meters.


     The gained uniform speed at the end of 10 seconds is 

        v = at = 1*10 = 10 m/s.


     The distance traveled uniformly is  

          d%5B2%5D = u*t = 10*20 = 200 meters.


     The deceleration value is  a%5Bdec%5D = v%2Ft = 10%2F5 = 2 m/s^2.  
     Formally, it is negative value; but for simplicity, I will use its modulus.

 
     The length of the decelerated part (or path) is

        d%5B3%5D = %28a%5Bdec%5D%2At%5E2%29%2F2 = %282%2A5%5E2%29%2F2 = 25 meters.


     Total distance traveled is  

        d = d%5B1%5D + d%5B2%5D + d%5B3%5D = 50 + 200 + 25 = 275 meters.



(b)  average speed is the total distance divided by the total elapsed time,  or

        275%2F%2810%2B20%2B5%29 = 275%2F35 = 7.857  m/s  (rounded).



(c)  Half the distance is  275%2F2 = 137.5 meters.

     It is longer then 50 meters (acceleration part) and shorter than 50+200 = 250.

     So, half of the distance is somewhere on the uniform part.

     Therefore, the time to get half the distance is  

         10 seconds PLUS time to cover 137.5-50 meters at the uniform rate of 10 m/s,  or

            the time to get half the distance = 10 + %28137.5-50%29%2F10 = 10 + 8.75 = 18.75 seconds.

Solved in full. All questions are answered.

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The designations are standard in Physics, so I do not decipher them on the way.

All formulas used are the standard formulas of Physics.