Question

A train is slowing down with an average acceleration of −5.0 m/s2 (negative sign shows deceleration). If its initial velocity is 50. m/s, how far does it travel in 6.0 seconds?

Answer 1

This problem requires 2 kinematics equations. What you first must do is find the Final Velocity with the given values... Vf = Vi + at ... which is Vf = 50 + (-5)(6) which would simplify to Vf = 30. Then you can plug this value along with the given values into the kinematics equation Vf^2=Vi^2 + 2ax. Plugged in this would be 30^2=50^2 + 2 (-5)x. This can then be simplified to find x (distance). Which would be x=160. This means that the train traveled 160m.

Answer 2

Answer:

the train travels 210 m

Explanation:

Kinematics equations of motion

$vf=vi+a*t$,   equation(1)

$vf^{2} =vi^{2} +2*a*d$ equation(2)

Vi= initial velocity (m/s)

Vf= final velocity (m/s)

a = acceleration $m/s^{2}$

d= traveling distance (m)

known information:

vi=50m/s

a=-5$m/s^{2}$

t=6 s

Calculation of the final speed replacing the information known in the equation 1

$vf=50-5*6$

$vf=50-30=20m/s$

Calculation of the distance replacing the information known  and the final velocity in the equation 2

$20^{2} =50^{2} -2*5*d\\400= 2500-10d\\10d=2500-400\\d=2100/d\\d=210m$