How to calculate spped and time?

Question

A train n a straight, level track has an initial speed of 35kn/hr. a uniform acceleration of 1.5 m/s^2 is applied while the train traels 200m. a)whaat is the speed of the train at the end of the distance?b) who long did it take the train to travel the 200 m?
someone please help i have been doing this one all day and i still cannot figure out what the answer is????

Answer 1

a) Use the following kinematic equation of motion:

2ax = [v_f]^2 - [v_0]^2, where:

a = acceleration
x = distance
v_f = final velocity
v_0 = initial velocity

Here, are solving for v_f:

v_f = sqrt(2ax + [v_0]^2)

Remember, to change km/hr to m/s for the initial velocity:

(35 km/hr)(1000 m / 1 km)(1 hr / 3600 sec) = 9.72 m/s

Plugging in our values, we get:

v_f = sqrt[(2)(1.5)(200) + (9.72)^2] = 26.35 m/s
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b) Use the following kinematic motion equation:

v_f = at + v_0

Solving for 't', you get --------> t = (v_f - v_0)/a

Plugging in our values, we get:

t = (26.35-9.72)/(1.5) ------> t = 11.09 seconds

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Hope this helps

Answer 2

a) Make an equation for speed in terms of time: speed = acceleration*time + initial speed

speed=35+1.5t

b) Make an equation with distance in terms of time, distance=initial speed*time + 1/2 acceleration*time squared

200=35t+.5*1.5(t^2)

Answer 3

woah...this is a tough problem...sorry i couldn't help ...just try searchin' online..best wishes...sorry for the inconvenience