A train starting from rest attains a velocity of 54 km per hour in 5 minutes assuming that the acceleration is uniform find the acceleration and the distance travelled by the train for attaining this velocity
Answers
Answer :-
→ Acceleration of the train = 0.05 m/s²
→ Distance travelled = 2250 m
Explanation :-
We have :-
• Initial velocity (u) = 0 m/s
• Final velocity (v) = 54 km/h
• Time taken (t) = 5 min = 300 sec
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Firstly, let's convert the unit of final velocity of the train (v) from "km/h" to "m/s" .
⇒ 1 km/h = 5/18 m/s
⇒ 54 km/h = 54(5/18)
⇒ 15 m/s
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By putting values in the 1st equation of motion, we would obtain acceleration of the train.
v = u + at
⇒ 15 = 0 + a(300)
⇒ 15 = 300a
⇒ a = 15/300
⇒ a = 0.05 m/s²
Now, we will calculate distance travelled by the train for attaining the final velocity by using the 3rd equation of motion .
v² - u² = 2as
⇒ (15)² - 0 = 2(0.05)s
⇒ 225 = 0.1s
⇒ s = 225/0.1
⇒ s = 2250 m
Provided that:
- Initial velocity = 0 mps
- Final velocity = 54 kmph
- Time = 5 minutes
To calculate:
- The acceleration
- Distance
Solution:
- The acceleration = 0.5 mps²
- Distance = 2250 m
Using concepts:
- Formula to convert kmph-mps
- Formula to convert min-sec
- Acceleration formula
- Third equation of motion
Using formulas:
Formula to convert kmph-mps:
Formula to convert min-sec:
Acceleration formula:
- We can use first equation of motion too for find acceleration.
First equation of motion:
Third equation of motion:
- We can use second equation of motion too for calculating the distance travelled.
Second equation of motion:
Where, v denotes final velocity, u denotes initial velocity, a denotes acceleration, t denotes time taken, s denotes displacement or distance or height.
Required solution:
✡️ Firstly let us convert kmph-mps!
✡️ Now let us convert min to sec!
✡️ Now let us find out the acceleration!
- We can find acceleration by using first equation of motion too!
✡️ Now let us find out the distance!
- We can use second equation of motion too for finding the distance travelled!