a train starts from rest and accelerates uniformly to achieve a velocity of 20m/sec in 10 sec.then train maintains this speed for next 200 sec.the breaks are then applied and train comes to rest in next 50 sec.calculate
a)acceleration in first 10 sec.
b)acceleration in last 50 sec.
c)the total distance traveled by the train in whole journey
d)average velocity of the trip.
a. (V-Vo)/t = (20-0)/10 = 2 m/s^2.
b. a=(V-Vo)/t = (0-20)/50 = -0.4 m/s^2.
c. d1 = 0.5a*t^2 = 1*10^2 = 100 m.
d2 = 20m/s * 200s = 4000 m.
d3 = Vo*t + 0.5a*t^2
d3 = 20*50 + (-0.2)*50^2 = 500 m.
d1+d2+d3 = 100 + 4000 + 500 = 4600 m. = Total dist. traveled.
d. T = t1 + t2 + t3 = 10 + 200 + 50 = 260 s.
Va = (d1+d2+d3)/T = 4600m / 260 = 17.7
m/s. = Average velocity.
To solve this problem, we will use the basic equations of motion. Let's go step by step:
a) To find the acceleration in the first 10 seconds, we can use the formula:
v = u + at
where:
v = final velocity = 20 m/s (given)
u = initial velocity = 0 m/s (train starts from rest, so the initial velocity is 0)
t = time = 10 s (given)
Rearranging the formula, we have:
a = (v - u) / t
Plugging in the values, we get:
a = (20 - 0) / 10 = 2 m/s^2
Therefore, the acceleration in the first 10 seconds is 2 m/s^2.
b) To find the acceleration in the last 50 seconds, we can assume that the train decelerates uniformly. We can use the same formula as above:
v = u + at
where:
v = final velocity = 0 m/s (train comes to rest, so the final velocity is 0)
u = initial velocity = 20 m/s (train maintains this speed for next 200 s, so the initial velocity is 20 m/s)
t = time = 50 s (given)
Rearranging the formula, we have:
a = (v - u) / t
Plugging in the values, we get:
a = (0 - 20) / 50 = -0.4 m/s^2
Therefore, the acceleration in the last 50 seconds is -0.4 m/s^2.
c) To calculate the total distance traveled by the train, we need to calculate the distance traveled during the acceleration phase and the distance traveled during the constant speed phase.
As the train starts from rest and accelerates uniformly, we can use the formula:
s = ut + 1/2 at^2
where:
s = distance
u = initial velocity = 0 (train starts from rest)
t = time
For the first 10 seconds (acceleration phase):
s1 = 0(10) + 1/2 (2)(10)^2 = 100 m
For the next 200 seconds (constant speed phase), the train maintains a speed of 20 m/s, so the distance traveled is:
s2 = 20(200) = 4000 m
For the last 50 seconds (deceleration phase):
s3 = 20(50) + 1/2 (-0.4)(50)^2 = 2500 m
Therefore, the total distance traveled by the train is:
Total distance = s1 + s2 + s3 = 100 + 4000 + 2500 = 6600 m
d) Average velocity is given by:
Average velocity = Total distance / Total time
Total time = 10 + 200 + 50 = 260 s.
Plugging in the values, we get:
Average velocity = 6600 / 260 = 25.38 m/s (rounded to two decimal places)
Therefore, the average velocity of the trip is approximately 25.38 m/s.