Physics Question (show working please)

Question

after stopping at a station, a train accelerates at a steady rate of 0.25ms^-2.

at what speed is the train moving after it has travelled 1.0 km?

Answer 1

2

Answer 2

The relation between distance, time and acceleration is

d = 0.5*a*t²

and velocty

v = a*t;

so t = v/a  d = 0.5*v²/a

v = √[2*a*d]

Answer 3

position as a function of time

a(t)=v'(t)=x''(t), so integrating a(t)=.25 twice, x(t)=.125t^2.

solve .125t^2=1000 to get the time corresponding to 1km displacement (remembering to watch your units), and plug in to the velocity functino (obtained by integrating acceleratino once).

if you aren't a calculus student, just memorize:

v(t)=at

x(t)=.5at^2

and use the above, ignoring the stuff about integrating

Answer 4

ewwww!!!! stop it!! i hate math. its like sooo hard.. ach... sorry... eww..

Answer 5

We can use the time-independent equation for this question:

v^2 = vi^2 + 2ad

Plugging in our numbers (after we convert the 1 km to meters):

v^2 = (0 m/s)^2 + 2(.25 m/s^2)(1000 m)

Solving for v:

v^2 = 0 + 500 m^2/s^2

v = 22.36 m/s

Answer 6

huh?

Answer 7

its stopped