A train accelerated from 10km/hr to 40km/hr in 2 minutes. How much distance does it
cover in this period? Assume that the tracks are straight?
Answers
Given :
• Initial velocity of the train = 10 km/hr
• Final velocity of the train = 40 km/hr
• Time = 2 minutes
To find :
• Distance covered by the trains in the time period of 2 minutes
Concept :
To find the distance covered by the train, firstly we need to find the acceleration. For that we will use the first equation of motion. Then to find the distance travelled use the second equation of motion. The resultant value after doing the required calculation will be the distance travelled by the train.
Equation of motion :
• First equation of motion :-
- v = u + at
• Second equation of motion :-
- S = ut + ½ at²
where,
• v denote the final velocity
• u denotes the initial velocity
• a denotes the acceleration
• S denotes the displacement/distance
• t denotes the timw
Solution :
Converting the initial and final velocity of the train from km/hr into m/s.
To convert any value from km/hr into m/s multiply it by 5/18.
Initial velocity in m/s :-
⠀⠀⠀⇒ Initial velocity = 10 × 5/18
⠀⠀⠀⇒ Initial velocity = 5 × 5/9
⠀⠀⠀⇒ Initial velocity = 25/9
⠀⠀⠀⇒ Initial velocity = 2.78
Initial velocity of the train = 25/9 m/s or 2.78 m/s
Final velocity in m/s :-
⠀⠀⠀⇒ Final velocity = 40 × 5/18
⠀⠀⠀⇒ Final velocity = 20 × 5/9
⠀⠀⠀⇒ Final velocity = 100/9
⠀⠀⠀⇒ Final velocity = 11.11
Final velocity of the train = 100/9 m/s or 11.11 m/s
Now, converting the time from minutes into seconds :-
As we know,
- 1 minute = 60 seconds
Mulitply the value by 60 to convert it from minutes into seconds.
⠀⠀⠀⇒ Time = 2 × 60
⠀⠀⠀⇒ Time = 120
Time = 120 seconds
Now, we will find the acceleration of the train.
Applying the first equation of motion :
⠀⠀⠀⇒ v = u + at
we have,
- v = 11.11 m/s
- u = 2.78 m/s
- t = 120 s
⠀⠀⠀⇒ Substitute the values :
⠀⠀⠀⇒ 11.11 = 2.78 + a(120)
⠀⠀⠀⇒ 11.11 = 2.78 + 120a
⠀⠀⠀⇒ 11.11 - 2.78 = 120a
⠀⠀⠀⇒ 8.33 = 120a
⠀⠀⠀⇒ 8.33/120 = a
⠀⠀⠀⇒ 0.069 = a
⠀⠀⠀⇒ On rounding off, we get :
⠀⠀⠀⇒ 0.07 = a
The acceleration of the train = 0.07 m/s²
Now, we can calculate the distance travelled.
Applying the second equation of motion :
⠀⠀⠀⇒ S = ut + ½ at²
⠀⠀⠀⇒ Substitute the values :
⠀⠀⠀⇒ S = (2.78)(120) + ½ (0.07)(120)(120)
⠀⠀⠀⇒ S = 333.6 + ½ (0.07)(120)(120)
⠀⠀⠀⇒ S = 333.6 + (0.07)(60)(120)
⠀⠀⠀⇒ S = 333.6 + 504
⠀⠀⠀⇒ S = 837.6
Therefore, the distance travelled by the train in the time period of 2 minutes = 837.60 m