Atrain pulls away from a station with a constant acceleration of 0.6 m/s 2 . a passenger arrives at a position on the track 4.5 s after the end of the train left that position. what is the slowest constant speed at which she can run and catch the train? answer in units of m/s.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

it has to be b, about 97 percent.

well clustered together.

points close on a scatter diagram