A freight train, traveling at a speed of 18.0 m/s, begins braking as it approaches a train yard. The train's acceleration while braking is –0.33 m/s2 . What is the train's speed after 23s?

v = 10.41 m/s

Step-by-step explanation:

using one of the equations of motion:

v = u + at

where v= final velocity

u= initial velocity

a = acceleration

t = time

u = 18.0 m/s

v = ?

a = –0.33 m/s2

t = 23s

v = 18.0 + (–0.33 x  23)

v = 18.0 + (-7.59)

v = 10.41 m/s

explanation:

step-by-step explanation:

one way is to plot 2 points and connect them which is what the answer choices seem to be doing

it seems to be using the y intercept and the slope

y=mx+b, m=slope, b=y-intercept (where the line hits the y axis)

slope=rise/run

y-intercept is where the line croses the y axis which is the y value where x=0

you seem to have given the wrong slope (you didn't put a number in front of te x, it should be 2/3 or 3/2 looking at the answer options)

y=? x-1

m=? because of your fault

b=-1

so it has a slope of ? and y intercept of -1, meaning it pases through (0,-1)

so draw point (0,-1) (so answer is one of the first 2 option)

if slope is 2/3, then go up 2 and right 3 then draw line through those 2 points

if slope is 3/2, then go up 3 and right 2 then draw line through those 2 points