The MRT train starting at an initial velocity of 30 kph travels a distance of 21km in 18 minutes. Determine the acceleration of the train at this instant in m/s^2.

Question

Illustrate an MRT train commencing a journey, moving at an initial speed of 30 kilometers per hour. The environment should depict a journey that covers a distance of 21 kilometers. To represent the concept of time passing, depict a large clock that shows 18 minutes have passed. Please make sure the image does not incorporate any explicit numerical values or text. Additionally, convey the idea of the train accelerating, perhaps through a keen use of motion blur in the background or vibrations emanating from the train itself.

The MRT train starting at an initial velocity of 30 kph travels a distance of 21km in 18 minutes. Determine the acceleration of the train at this instant in m/s^2.

Answer 1

vf^2 = vo^2 + 2a*distance change 18 min to seconds, km to meters 30kph to m/s

and solve for a.

Answer 2

0.0865 m/s˄2

Answer 3

To determine the acceleration of the train, we can use the equation of motion:

\(\text{Acceleration} (a) = \frac{{\text{Change in velocity} (v)}}{{\text{Change in time} (t)}}\)

Given:
Initial velocity (\(u\)) = 30 kph
Distance (\(s\)) = 21 km
Change in time (\(t\)) = 18 minutes

First, let's convert the initial velocity and distance to meters and change the time to seconds for consistent units.

Initial velocity (\(u\)) = 30 kph = 30 × \(\frac{1000}{3600}\) m/s = 8.33 m/s (approximately)
Distance (\(s\)) = 21 km = 21 × 1000 m = 21000 m
Change in time (\(t\)) = 18 minutes = 18 × 60 s = 1080 s

Next, we can calculate the change in velocity (\(v\)) using the formula:

\( v = \frac{s}{t} \)

\( v = \frac{21000}{1080} \) m/s
\( v = 19.44 \) m/s (approximately)

Finally, substituting the values into the formula for acceleration:

\( a = \frac{v - u}{t} \)
\( a = \frac{19.44 - 8.33}{1080} \) m/s²
\( a = \frac{11.11}{1080} \) m/s²
\( a \approx 0.01028 \) m/s² (approximately)

Therefore, the acceleration of the train at this instant is approximately 0.01028 m/s².

Answer 4

To determine the acceleration of the train, we need to use the formula:

acceleration = change in velocity / time

First, let's convert the initial velocity from kph to m/s:
30 kph = 30 * (1000/3600) m/s = 8.33 m/s

Next, we need to determine the final velocity. We know that the train traveled a distance of 21 km in 18 minutes. Let's convert that to meters and seconds for consistent units:
21 km = 21000 m
18 minutes = 18 * 60 seconds = 1080 seconds

To find the final velocity, we can use the formula:
final_velocity = initial_velocity + (acceleration * time)

In this case, the initial velocity is 8.33 m/s, and the time is 1080 seconds. We want to solve for the final velocity, so rearranging the formula gives us:
final_velocity = initial_velocity + (acceleration * time)
final_velocity = 8.33 m/s + (acceleration * 1080 seconds)

Since the train starts at rest, the final velocity is the velocity at this instant. Therefore, we have:
8.33 m/s + (acceleration * 1080 seconds) = final_velocity

Now, let's use the formula to find the acceleration:
acceleration = (final_velocity - initial_velocity) / time
acceleration = (final_velocity - 8.33 m/s) / 1080 seconds

We are given that the train traveled a distance of 21 km, which means it came to a stop at the end. When the train comes to a stop, the final velocity is zero. Therefore, we have:
(0 m/s - 8.33 m/s) / 1080 seconds = acceleration

Calculating this expression, we find:
acceleration = -8.33 / 1080 m/s^2

Therefore, the acceleration of the train at this instant is approximately -0.0077 m/s^2. The negative sign indicates that the train is decelerating.