The greatest accelaration or deaccelaration that a train may have is a . The minimum time in which the train can get from one situation to the next at distance a.
Answers
Answer:
The greatest accelaration or deaccelaration that a train may have is a . The minimum time in which the train can get from one situation to the next at distance a.
Question :
The greatest accelaration or deaccelaration that a train may have is a . The minimum time in which the train can get from one situation to the next at distance a.
Solution :
Note : To solve the numerical based on equation of motion we should remember the few points :
> If the body is at rest then we should take the initial velocity as zero.
> If the body stops after sometime then final velocity is taken as zero.
> If the body moves with uniform velocity then acceleration is taken as zero.
There is no limit on the speed of the train, so minimum time will be when the train will accelerate for t/2 and deaccelerate for t/2. So time taken will calculated by s = it + 1/2at² , where u = 0 , displacement = s/2 , time = T/2 , same distance will be covered while acceleration as well as deacceleration.
So T = 2√s/a
Alternate ,
Distance is equal to the Area under the velocity - time graph on the axis of time.
Let us consider the total distance travelled by train is taken as d. It covers the first half with an acceleration and the second half it decelerates therefore it takes half time for both the cases.
Train is initially at rest therefore the initial velocity u = 0.
By using Area of triangle,
Distance = 1/2 × base × height
D = 1/2 × t × 1/2at
D = 1/4at²
T² = 4d/a
T = √4d/a
t = 2√d/a