A train on a straight, level track has an initial speed of 45 km/h. a uniform acceleration of 1.50 m/s^2 is?

Question

applied while the train travels 200 m

a. what is the speed of the train at the end of this distance?

b. how long did it take for the trian to travel the 200m?

Answer 1

<< what is the speed of the train at the end of this distance?  >>

Your working equation is

Vf^2 - Vo^2 = 2as

where

Vf = final velocity

Vo = initial velocity = 45 km/hr. = 12.5 m/sec (given)

a = acceleration = 1.5 m/sec^2

s = distance travelled = 200 meters

Substituting appropriate values,

Vf^2 - 12.5^2 = 2(1.5)(200)

Vf^2 = 12.5^2 + 2(1.5)(200)

Solving for Vf,

Vf = 27.5 m/sec = 99 km/hr.

<< how long did it take for the trian to travel the 200m? >>

There are several options on how to solve for this travel time of the train but the easiest way is to use the formula

a = (Vf - Vo)/T

where

T = time for train to travel 200 meters

and all the other terms have been previously defined

Substituting values,

1.50 = (27.5 - 12.5)/T

Solving for T,

T = 15/1.5

T = 10 seconds

Answer 2

a. Convert km/hr into m/s by multiplying by (1000 / 3600).

vf = (vi² + 2ad)^(1/2)

= ((12.5 m/s)² + (2 x 1.50 m/s² x 200 m))^(1/2)

= 27.5 m/s

b. t = ∆v / a

= (27.5 m/s - 12.5 m/s) / 1.50 m/s²

= 10.0 s