Q29PE - College Physics (Urone)

Question

Freight trains can produce only relatively small accelerations and decelerations. 

(a) What is the final velocity of a freight train that accelerates at a rate of \({\bf{0}}.{\bf{0500}}{\rm{ }}{\bf{m}}/{{\bf{s}}^{\bf{2}}}\) for \({\bf{8}}.{\bf{00}}{\rm{ }}{\bf{min}}\), starting with an initial velocity of\({\bf{4}}.{\bf{00}}{\rm{ }}{\bf{m}}/{\bf{s}}\)? 

(b) If the train can slow down at a rate of \({\bf{0}}.{\bf{0500}}{\rm{ }}{\bf{m}}/{{\bf{s}}^{\bf{2}}}\), how long will it take to come to a stop from this velocity? 

(c) How far will it travel in each case?

Step-by-Step Solution

Answer

1. \({\bf{28}}\;{\rm{m/s}}\) .
2. \({\bf{50}}.{\bf{909}}\;{\bf{s}}\).
3. \({\bf{7680}}\;{\bf{m}}\). And \({\bf{712}}.{\bf{727}}\;{\bf{m}}\).

Step 1: Given data

List of known values:

• Time =\({\bf{8}}\;{\bf{min}}{\rm{ }} = {\rm{ }}{\bf{8}} \times {\bf{60}}\;{\rm{s }} = {\rm{ }}{\bf{480}}\;{\bf{s}}\).
• Initial velocity U = \({\bf{4}}.{\bf{00}}\;{\rm{m/s}}\).
• Acceleration a = \({\bf{0}}.{\bf{0500}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).

Step 2: Final velocity of the train

a)

We can use the equation of motion as:

\(V = U + at\)

Where V is the final velocity, U is the initial velocity, a is acceleration, and t is the time taken.

Substituting values in the above expression, we get,

\(\begin{array}{l}V = 4 + (0.05) \times (480)\\V = 28\;{\rm{m/s}}\end{array}\) 

Hence the final velocity of the train is \({\bf{28}}\;{\rm{m/s}}\).

Step 3: Time is taken to stop the train

b)

List of known values:

• Initial velocity U= \({\bf{28}}\;{\rm{m/s}}\).
• Acceleration a = \( - {\bf{0}}.{\bf{55}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).
• Final velocity V = \({\bf{0}}\;{\rm{m/s}}\).

We can use the equation of motion as:

\(V = U + at\)

Where V is the final velocity, U is the initial velocity, a is acceleration, and t is the time taken.

Substituting values in the above expression, we get,

\(\)\(\begin{array}{l}0 = 28 + ( - 0.55) \times (t)\\t = \frac{{28}}{{0.55}}\\t = 50.909\;{\rm{s}}\end{array}\)

Hence the time taken by the train to stop is \({\bf{50}}.{\bf{909}}\;{\bf{s}}\).

Step 4: The distance traveled by train in both cases

c) 

In the above case, the train will cover the distance d as:

\(d = ut + \frac{1}{2}a{t^2}\)

Where d is the distance traveled, u is the initial velocity, a is acceleration, and t is the time taken.

Substituting values in the above expression, we get,

\(\begin{array}{l}d = 4(480) + \frac{1}{2}(0.05) \times {(480)^2}\\d = 7680\;{\rm{m}}\end{array}\) 

Hence the distance traveled by train is \({\bf{7680}}\;{\bf{m}}\).

In the above case, the train will cover the distance d as:

\(d = ut + \frac{1}{2}a{t^2}\)

Where d is the distance traveled, u is the initial velocity, a is acceleration, and t is the time taken.

Substituting values in the above expression, we get,

\(\begin{array}{l}d = 28 \times (50.909) + \frac{1}{2}( - 0.55) \times {(50.909)^2}\\d = 712.727\;{\rm{m}}\end{array}\)

Hence the distance traveled by train is \({\bf{712}}.{\bf{727}}\;{\bf{m}}\).