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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
= = = = 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
= u*t = 10*20 = 200 meters.
The deceleration value is = = = 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
= = = 25 meters.
Total distance traveled is
d = + + = 50 + 200 + 25 = 275 meters.
(b) average speed is the total distance divided by the total elapsed time, or
= = 7.857 m/s (rounded).
(c) Half the distance is = 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 + = 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.