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

Answer:

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$m/s^{2}$, t=14s

v=0+(1.6*14)

v=22.4m/s

Distance covered in the first lap

$s=ut+\frac{1}{2}at^{2}$

$s=0+\frac{1}{2}1.6*14^{2}$

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=?

$a=\frac{v-u}{t}\\\ at=v-u\\\\ t=\frac{v-u}{a}$

$t=\frac{0-22.4}{-3.5}$

t=6.4s

distance covered in the last lap

$S=ut+\frac{1}{2}at^{2}$

$S=22.4*6.4+\frac{1}{2}*-3.5*6.4^{2}$

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