The length ofa train is 44.5 m. its front is 100. m from a pole. it accelerates from rest at 0.500 m/s^2. (a) how long does it take to go past the pole? (b) at what speeds do its front and rear pass the pole?

a) It takes the train 24.0 s to go past the pole.

b) The rear passes the pole at 12 m/s

The front passes the pole at 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line with constant acceleration are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0  + a · t

Where:

x = position at time t.

x0 = initial positon.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity at time t.

a) Let´s place the origin of the frame of reference at the pole. At t = 0, the front of the train is at -100 m and the rear is at -144.5 m. Then, when the whole train passes the pole, the front will be at 44.5 m.

Let´s find the time it takes the front of the train to reach that position:

x = x0 + v0 · t + 1/2 · a · t²

44.5 m = -100 m + 0 m/s · t + 1/2 · 0.500 m/s² · t²

(44.5 m + 100 m) / (1/2 · 0.500 m/s²) = t²

t = 24.0 s

It takes the train 24.0 s to go past the pole.

b) The rear passes the pole at t = 24.0 s, then, using the equation for velocity:

v = v0  + a · t

v = 0 m/s + 0.500 m/s² · 24.0 s = 12 m/s

The rear passes the pole at 12 m/s

Now, we have to find how much time it takes the front to pass the pole. Using the equation of position, we have to calculate the time at which the position of the front is 0 m:

x = x0 + v0 · t + 1/2 · a · t²

0 m = -100 m + 0 m/s · t + 1/2 · 0.500 m/s² · t²

2 · 100 m / 0.500 m/s² = t²

t = 20.0 s

The velocity of the front at that time will be:

v = v0  + a · t

v = 0 m/s + 0.500 m/s² · 20.0s

v = 10 m/s

The front passes the pole at 10 m/s

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form