Been trying to solve this problem for the loooooongest time. Any help would be REALLY REALLY appreciated. THANKS!
A subway train starts from rest at a station and accelerates at a rate of 1.60 for 14.0 . It runs at constant speed for 69.3 and slows down at a rate of 3.49 until it stops at the next station. Find the total distance covered.
A subway train starts from rest at a station and accelerates at a rate of 1.60 for 14.0 . It runs at constant speed for 69.3 and slows down at a rate of 3.49 until it stops at the next station. Find the total distance covered.
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as it takes 14 seconds to accelerate at 1.60 m/s^2
it must take 14 * 1.6 /3.49 = 6.42 secs to slow down again.
Now what speed did it reach?
v = at = 1.6 * 14 = 22.4 m/s
During acceleration and deceleration the AVERAGE speed is 1/2 max speed = 11.2 m/s
So to get the total distance we have distance = sum ( speed * time)
= 11.2 * 14 + 22.4 * 69.3 + 11.2 * 6.42
= 1.78 Km
it must take 14 * 1.6 /3.49 = 6.42 secs to slow down again.
Now what speed did it reach?
v = at = 1.6 * 14 = 22.4 m/s
During acceleration and deceleration the AVERAGE speed is 1/2 max speed = 11.2 m/s
So to get the total distance we have distance = sum ( speed * time)
= 11.2 * 14 + 22.4 * 69.3 + 11.2 * 6.42
= 1.78 Km