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) (Velocity change)/Time = 2.0 m/s^2
b) (Velocity change)/Time = -0.4 m/s^2
c) X1 + X2 + X3, where X1 is the distance travelled accelerating; X2 is the distance travelled at constant speed and X3 is the distance travelled while decelerating.
= 10*10 + 20*200 + 10*50 = ___ m
Average speed was used for X1 and X3.
d) Divide previous answer by total elapsed time, 260 s.
To solve this problem, we need to use the equations of motion for uniformly accelerated motion. Let's break down each part of the question:
a) Acceleration in the first 10 seconds:
We know the final velocity (v) is 20 m/s, the initial velocity (u) is 0 m/s, and the time (t) is 10 seconds. The formula to find acceleration (a) is:
a = (v - u) / t
Plugging in the values, we get:
a = (20 - 0) / 10
a = 2 m/s²
b) Acceleration in the last 50 seconds:
Since the train applies brakes and comes to a rest, the final velocity (v) is 0 m/s, the initial velocity (u) is 20 m/s, and the time (t) is 50 seconds. Using the same formula, we have:
a = (v - u) / t
a = (0 - 20) / 50
a = -0.4 m/s²
c) Total distance traveled by the train in the whole journey:
To find the total distance, we need to calculate the distance the train travelled during the first 10 seconds (with constant acceleration), the distance it covered during the next 200 seconds (with constant velocity), and the distance it covered in the last 50 seconds (with deceleration).
The distance covered in the first 10 seconds (d1) can be calculated using the formula:
d1 = ut + (1/2) * a * t²
Plugging in the values, we get:
d1 = 0 * 10 + (1/2) * 2 * (10)²
d1 = 100 m
The distance covered in the next 200 seconds is simply:
d2 = v * t
d2 = 20 * 200
d2 = 4000 m
The distance covered in the last 50 seconds can be calculated using the same formula as before:
d3 = ut + (1/2) * a * t²
Plugging in the values, we get:
d3 = 20 * 50 + (1/2) * (-0.4) * (50)²
d3 = 1000 - 250 m
d3 = 750 m
The total distance traveled is the sum of all three distances:
Total distance = d1 + d2 + d3
Total distance = 100 + 4000 + 750
Total distance = 4850 m
d) Average velocity of the trip:
The average velocity (v_avg) is given by the total distance divided by the total time. We have already calculated the total distance, and the total time is 10 + 200 + 50 = 260 seconds.
v_avg = Total distance / Total time
v_avg = 4850 / 260
v_avg ≈ 18.65 m/s
So, the answers to the questions are:
a) Acceleration in the first 10 sec: 2 m/s²
b) Acceleration in the last 50 sec: -0.4 m/s²
c) Total distance traveled by the train: 4850 m
d) Average velocity of the trip: 18.65 m/s