Question 1207778
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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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<pre>
(a)  To find the length of the accelerated part (or path), use the formula

        {{{d[1]}}} = {{{(a[acc]t^2)/2}}} = {{{(1*10^2)/2}}} = {{{100/2}}} = 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[2]}}} = u*t = 10*20 = 200 meters.


     The deceleration value is  {{{a[dec]}}} = {{{v/t}}} = {{{10/5}}} = 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[3]}}} = {{{(a[dec]*t^2)/2}}} = {{{(2*5^2)/2}}} = 25 meters.


     Total distance traveled is  

        d = {{{d[1]}}} + {{{d[2]}}} + {{{d[3]}}} = 50 + 200 + 25 = 275 meters.



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

        {{{275/(10+20+5)}}} = {{{275/35}}} = 7.857  m/s  (rounded).



(c)  Half the distance is  {{{275/2}}} = 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 + {{{(137.5-50)/10}}} = 10 + 8.75 = 18.75 seconds.
</pre>

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.