Distance

[Guest]

A train starts at a point distant A feet from the terminal, and travels with an initial velocity V ft. per sec., and acceleration of F ft. per sec. per sec., and is traveling for a time T sec. Determine distance from the following:

A = 12 ft.
V = 20 ft. per sec.
F = 30 ft. per sec. per sec.
T = 25 sec.

Answer provided with problem: 9,887 ft.

How do I solve ?

[Guest]

S = distance travelled
A = initial position
V = initial velocity
F = acceleration ft per sec. per sec.
T = time taken
S = A + VT + 1/2*F*t^2

[Guest]

" S = distance travelled
A = initial position
V = initial velocity
F = acceleration ft per sec. per sec.
T = time taken
S = A + VT + 1/2*F*t^2 "

S = 12 + (20 x25) + (1/2 x 30 x 25 x 25)

S = 12 + 500 + 9375

S = 9887 ft.

Where am I correct ?

Also, why is F needed ?

Could rate x time = distance be sufficient ?

Thanks.

[Guest]

Yes, looks correct.

If the train was moving at a constant velocity the acceleration would be Zero and F = 0. So F would not be needed

Rate x Time = Distance is only for constant rate. Rate is same as speed or velocity

The total distance in this question is the initial position (A) + initial velocity x time (V*T) + distance travelled due to acceleration (1/2*F*T^2)

[Guest]

Ok, thanks.

[Guest]

Just out of interest, I wanted to find rate:

9887 / 5280 = 1.87 mi.

25 / 3600 = .0069 hrs.

1.87 / .0069 = 271 mph.

Where am I correct ?

[Guest]

In the original question I assumed the train moved 12 feet from the terminal (we don't know speed or time) to point A and at point A it had an initial velocity of 20 feet per second and an acceleration of 30 feet per sec. per sec. and at point A T was Zero, T = 0.
The distance S we calculated was the distance from the actual terminal, not from point A.
So the train moved 9887 - 12 = 9875 feet from point A during the time T = 25 seconds.
The train could have stopped overnight between the terminal and point A, we do not know, all we know it was moving past point A with an initial velocity of 20 feet per sec. and was accelerating at 30 feet per sec. per sec. and we assume constant acceleration of 30 feet per sec per sec. during all of the 25 second period.

We can work out the equivalent average rate or speed during this time using the formula.... Rate = Distance / Time. This will be the rate or speed the train would have needed to be travelling at, constantly throughout, to cover the same distance in 25 seconds. Travelling at constant velocity.

But in the question the train was travelling at constant acceleration.... that means it was speeding up from its initial velocity as it travelled.

So....... Rate = Distance / Time

9875 / 5280 = 1.87 mi.
25 / 3600 = 0.0069 hrs

Rate = 1.87 / 0.0069 = 271 mph. constant speed throughout......
......same as your answer......

In the original question the train passed point A at 20 feet per second
So in first second this speed converted to mph is ..... (20 / 5280) x 3600 = 13 .64 mph
And it accelerated up from this speed to 20 + (30 x 25) = 770 feet per second at the end of 25 seconds.
This is equivalent to (770 / 5280) x 3600 = 525 mph in the 25th second.

[Guest]

Ok, thank you again.

[leesajohnson]

S = A + VT + 1/2*F*t^2

S= 12 + (20 x25) + (1/2 x 30 x 25 x 25)

S = 9887 ft.

[Guest]

Thanks Ms. Johnson.