A train accelerates from 36kmph to 72kmph on covering a distance of 50m. calculate the acceleration of the train and the time to cover the distance 50m
Answers
Answer:3.3s
Explanation:
u=36km/h=10m/s
v=72km/h=20m/s
s=50m
v^2-u^2=2as
20^2-10^2=2*a*50
a=300/
Answer:
Acceleration = 3 m/s²
Time taken = 3.33 s
Explanation:
Given:
- Initial velocity = 36 km/hr = 10 m/s
- Final velocity = 72 km/hr = 20 m/s
- Distance covered = 50 m
To Find:
- Acceleration of the train
- Time taken
Solution:
First we have to find the acceleration of the train.
By the third equation of motion,
v² - u² = 2as
where v = final velocity
u = initial velocity
a = acceleration
s = distance
Substituting the data,
20² - 10² = 2 × a × 50
400 - 100 = 100 a
100a = 300
a = 300/100
a = 3 m/s²
Hence acceleration of the train is 3 m/s².
Now we have to find the time taken.
By the first equation of motion,
v = u + at
where v = final velocity
u = initial velocity
a = acceleration
t = time
Substitute the data,
20 = 10 + 3t
20 - 10 = 3t
3t = 10
t = 10/3
t = 3.33 s
Hence the time taken by the train is 3.33 s.
Notes:
The three equations of motion are:
- v = u + at
- s = ut + 1/2 × a × t²
- v² - u² = 2as