A train moving at an initial speed of 40m/sec puts on its brakes producing a deceleration of 0.8m/sec^2.?

Question

a) How long will it take the train to travel the next 110m?

b) At what speed will it be travelling at the end of this 110m?

Answer 1

WHO CARES TAKE PLANE, TRAINS ARE SO YESTERDAY

Answer 2

50 sec

Answer 3

x=distance traveled = 110m

t=time

v0=initial velocity=40m/s

a=acceleration speed, -0.8 m/s^2

The general formula for distance traveled with constant acceleration is:

x=v0*t+1/2a*t^2

it's derivative, gives the speed as a function of time:

v=v0+a*t

we can determine t from the 1st formula by using the quadratic formula.

x=v0*t+1/2a*t^2

1/2a*t^2+v0*t-x=0

t=(-v0 +/- sqrt(v0^2-4*1/2*a*-x)) / a

t=(-v0 +/- sqrt(v0^2+2*a*x)) / a

Choose the positive t. Then enter this t into the second formula to get the velocity at that timepoint.

Answer 4

There are three equations of motion.

v=u+at............equation1

v^2=u^2+2as........equation2

s=ut + 1/2 at^2.......equation3

where v is the final speed, u is the starting speed, t is time and s is the distance.

(b) Find v from equation 1

(a) Put that value into equation2 to find the distance, s.

Answer 5

Vo = 40m/s

A = -0.8m/s^2

X = 110m

Where Vo = intial velocity

A = Acceleration

X = Displacement

T = Time

Form of equation:

a) X = VoT + (1/2)AT^2

=>110 = 40(T) + (1/2)(-0.8)T^2

=>0.4T^2 - 40T + 110 = 0

=> 4T^2 - 400T + 1100 = 0

=> T^2 - 100T + 275 = 0

=> T = [100+-sqrt(100^2 - 4(275)) ] / 2

Slove for T...where T>0

b) Form of Equation:

V^2 = Vo^2 + 2AX

=>V^2 = 40^2 + 2(-0.8)(110)

Solve for V..

Answer 6

do your own homework

Answer 7

Im 19 and i cant even pass Algebra 1 dude haha, if you want help on writing, history, or astrology im your man lolz.