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Q2:
To do this you divide the journey into 3 pieces. A: The acceleration; B: The constant speed and C: The deceleration.
A: You know u = 0; it starts at rest; v = 20 - given; t is unknown; s is unknown. a = 1.25. with these variables you can calculate t.
v = u + at = 20 = 0 + 1.25 x t. Therefore t = 20/1.25 = 16 seconds.
To compute distance, apply s = ut + 1/2 at^2. here all variables are now known except for s.
s = 0 x 16 + 1/2 x 1.25 x 16^2 = 0 + 1/2 x 1.25 x 256 = 160m
B: the distance here is given - 1560m
C: here you know u is 20ms-1, v = 0, a = -2 and t is unknown. You use the same equation to find t as per part A.
v = u + at = 0 = 20 x -2t so t = 20/2 = 10 seconds.
From there plug that into the equation s = ut +1/2 at^2 to find distance.
s = 20 x 10 + 1/2 x -2 x 10^2 = 200 - 100 = 100m
Then you add the three components together: A + B + C = 160 + 1560 + 100 = 1820m
part (b)
Total time of the journey is the time for each component added together. To calculate the distance we have computed time for A and C. Now compute the time it takes to do the constant speed:
s = ut + 1/2at^2. since it is constant speed a = 0 so s = ut. Since s is given at 1560m and u is given at 20ms-1 then t = 1560 / 20 = 78 seconds.
Total time is therefore: 16 + 78 + 10 = 104 seconds.
Part (c)
Average speed of the journey is simply distance travelled divided by time. You have the total distance travelled at 1820m (part a) and total time of 104 seconds. Therefore average speed is:
1820 / 104 = 17.5ms-1