First of all, make the conversion
72 km/h = 20 m/s
After accelerating 60 seconds at 0.75 m/s^2, the speed increases by 45 m/s to 65 m/s. The average speed during that minute is 105/2 = 52.5 m/s, and the distance travelled is 52.5 m/s * 60 s = 3150 m.
After one more minute (60 seconds) of decelerating at -1.0 m/s^2, the speed drops to
105 - 60 = 45 m/s. The average speed during the second minute is (105 + 45)/2 = 75 m/s. The distance traveled during the second minute is 75*60 = 4500 m.
Add the distances traveled during the first and second minutes for the final answer.
You could also integrate the V(t) function for two minutes. The answer should be the same.
A train passing at point A at a speed of 72 kph accelerates at .75 m/s2 for one minute along a straight path then decelerated at 1.0 m/s2. How far in km from point A will it be 2 minutes after passing point A?
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