Question

A subway 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.

Answer 1

1796.48m

Step-by-step 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
`m/s^(2)`, t=14s

v=0+(1.6*14)

v=22.4m/s

Distance covered in the first lap

`s=ut+(1)/(2)at^(2)`

`s=0+(1)/(2)1.6*14^(2)`

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
`m/s^(2)`, t=?

`a=(v-u)/(t)\\\ at=v-u\\\\ t=(v-u)/(a)`

`t=(0-22.4)/(-3.5)`

t=6.4s

distance covered in the last lap

`S=ut+(1)/(2)at^(2)`

`S=22.4*6.4+(1)/(2)*-3.5*6.4^(2)`

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