Asubway train starts from rest at a station and accelerates at a rate of 1.60m/s2 for 14.0 s . it runs at constant speed for 70.0 s and slows down at a rate of 3.50m/s2 until it stops at the next station. find the total distance covered.

1796.48m

Explanation:

sign notation; v= final velocity,u=initial velocity , a=acceleration,  S= distance , t=time

to find the velocity during acceleration(v) we use the formula.

v=u+at

u=0,a=1.6, t=14s

v=0+(1.6*14)

v=22.4m/s

Distance covered in the first lap

s=156.8m

In the second lap,distance covered

S=vt       ,t=70s

S=22.4*70

S=1568m

it the last lap, u=22.4, v=0, a=-3.5, t=?

t=6.4s

distance covered in the last lap

S=71.68m

Total distance covered is equal to sum of distance covered in the three laps.

Total distance=156.8+1568+71.68

               =1796.48m

the answer is to check all of them! they all apply well at least i think so!

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